{"id":9204,"date":"2019-11-30T12:27:40","date_gmt":"2019-11-30T06:57:40","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=9204"},"modified":"2021-07-02T15:44:23","modified_gmt":"2021-07-02T10:14:23","slug":"1st-puc-physics-question-bank-chapter-14","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-physics-question-bank-chapter-14\/","title":{"rendered":"1st PUC Physics Question Bank Chapter 14 Oscillations"},"content":{"rendered":"

You can Download Chapter 14 Oscillations Questions and Answers, Notes, 1st PUC Physics Question Bank with Answers<\/a> Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 1st PUC Physics Question Bank Chapter 14 Oscillations<\/h2>\n

1st PUC Physics Oscillations Textbook Questions and Answers<\/h3>\n

Question 1.
\nWhich of the following examples represent periodic motion?
\n(a) A swimmer completing one (return) trip from one bank of a river to the other and back.
\n(b) A freely suspended bar magnet displaced from its N-S direction and released.
\n(c) A hydrogen molecule rotating about its center of mass.
\n(d) An arrow released from a bow.
\nAnswer:
\n(b) and (c)<\/p>\n

Question 2.
\nWhich of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion?<\/p>\n

    \n
  1. the rotation of earth about its axis.<\/li>\n
  2. motion of an oscillating mercury column in a U-tube.<\/li>\n
  3. motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lowermost point.<\/li>\n
  4. general vibrations of a poly-atomic molecule about Its equilibrium position.<\/li>\n<\/ol>\n

    Answer:<\/p>\n

      \n
    1. Periodic but not SHM (Simple Harmonic Motion)<\/li>\n
    2. SHM<\/li>\n
    3. SHM<\/li>\n
    4. Periodic but not SHM<\/li>\n<\/ol>\n

      \"KSEEB<\/p>\n

      Question 3.
      \nFigure below depicts four x<\/p>\n

       <\/p>\n

       <\/p>\n

       <\/p>\n

      -t plots for linear motion of a particle. Which of the piots represent periodic motion? What is the period of motion (in case of periodic motion)?
      \n\"1st
      \n\"1st
      \nAnswer:
      \n(b) and (d) are in periodic motion with period of 2 sec.<\/p>\n

      Question 4.
      \nWhich of the following functions of time represent
      \n(a) simple harmonic,
      \n(b) periodic but not simple harmonic, and
      \n(c) non-periodic motion?
      \nGive period for each case of periodic motion (\u03c9 is any positive constant):
      \n(a) sin \u03c9 t – cos \u03c9 t
      \n(b) sin3<\/sup> \u03c9 t
      \n(c) 3 cos( \u03c0\/4 – 2 \u03c9 t)
      \n(d) cos \u03c9 t + cos 3 \u03c9 t + cos 5 \u03c9 t
      \n(e) exp (- \u03c92<\/sup> t2<\/sup>)
      \n(f) 1 + \u03c9 t+ \u03c92<\/sup> t2<\/sup>
      \nAnswer:
      \n(a) sin \u03c9 t – cos \u03c9 t = \\(\\sqrt{2}\\left(\\frac{1}{\\sqrt{2}} \\sin \\omega t-\\frac{1}{\\sqrt{2}} \\cos \\omega t\\right)\\)
      \n= \\(\\sqrt{2} \\sin (\\omega t-\\pi \/ 4)\\) ; SHM , period = \\(\\frac{2 \\pi}{\\omega}\\)
      \n(b) sin3<\/sup> \u03c9 t = \\(\\frac{1}{4}\\) (3 sin \u03c9t – sin 3\u03c9t) both terms are in SHM hence sin3<\/sup> \u03c9t is periodic.
      \nperiod = \\(\\frac{2 \\pi}{\\omega}\\)
      \n(c) 3 cos (\u03c0\/4 – 2 \u03c9 t) S H M with period = \\(\\frac{\\pi}{\\omega}\\)
      \n(d) cos \u03c9 t + cos 3 \u03c9 t + cos 5 \u03c9 t Superposition of 3 periodic motion,period = \\(\\frac{2 \\pi}{\\omega}\\)
      \n(e) exp (- \u03c92<\/sup> t2<\/sup>) non periodic motion
      \n(f) 1 + \u03c9 t+ \u03c92<\/sup> t2<\/sup> non periodic motion<\/p>\n

      Question 5.
      \nA particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration, and force on the particle when it is
      \n(a) at the end A,
      \n(b) at the end B,
      \n(c) at the mid-point of AB going towards A,
      \n(d) at 2 cm away from B going towards A,
      \n(e) at 3 cm away from A going towards B, and
      \n(f) at 4 cm away from B going towards A.
      \nAnswer:
      \n\"1st<\/p>\n

      Question 6.
      \nWhich of the following relationships between the acceleration \u03b1 and the displacement x of a particle involve simple harmonic motion?<\/p>\n

        \n
      1. \u03b1 = 0.7x<\/li>\n
      2. \u03b1 = -200x2<\/sup><\/li>\n
      3. \u03b1 = -10x<\/li>\n
      4. \u03b1 = 100x3<\/sup><\/li>\n<\/ol>\n

        Answer:
        \n\u03b1 = -10 x (\u2235 acceleration is proportional and opposite to displacement)<\/p>\n

        \"KSEEB<\/p>\n

        Question 7.
        \nThe motion of a particle executing simple harmonic motion is described by the displacement function, x(t) = A cos (\u03c9 t + \u03a6). If the initial (t = 0) position of the particle is 1 cm and Its initial velocity is \u03c9 cm\/s, what are its amplitude and initial phase angle? The angular frequency of the particle is \u03c0 s-1<\/sup>. If instead of the cosine function, we choose the sine function to describe the SHM: x = B sin (\u03c9 t + \u03b1), what are the amplitude and initial phase of the particle with the above initial conditions.
        \nAnswer:
        \nGiven: x(t) = A cos(\u03c9 t + \u03a6) ……. (1)
        \nat t = 0, x(0) = 1cm, \u03c9 = \u03c0 s-1<\/sup> ,
        \nsubstituting in (1) we get
        \n1 cm = A cos (0 \u00d7 \u03c0 + \u03a6)
        \n\u21d2 A cos \u03a6 =1\u00a0 \u00a0 \u00a0…… (2)
        \nDifferentiate equation (1) w.r.t. t
        \n\u21d2 \\(\\frac{\\mathrm{d} \\mathrm{x}}{\\mathrm{dt}}\\) – A \u03c9 sin (\u03c9 t + \u03a6)\u00a0 \u00a0 …… (3)
        \nat t = 0, v = \u03c9 cm \/s, \u03c9 = \u03c0 s-1<\/sup>
        \nSubstituting in 3
        \n\u21d2 \u03c9 = – A \u03c9sin \u03a6
        \n\u21d2 1 = – Asin \u03a6\u00a0 \u00a0 \u00a0….. (4)
        \nDivide equation (4) by (2)
        \n\u21d2 1 = \\(\\frac{-\\sin \\phi}{\\cos \\phi}\\)
        \n\u21d2 tan \u03a6 = -1
        \n\u21d2 \u03a6 = \\(\\frac{-\\pi}{4}\\)
        \n\u03a6 = initial phase.
        \nSubstitute \u03a6 value in equation (2) we get
        \n1 = A cos \\(\\left(\\frac{-\\pi}{4}\\right)\\)
        \n\u21d2 A = \\(\\sqrt{2} \\mathrm{cm}\\)
        \nIf x (t) = B sin (\u03c9 t + \u03b1)\u00a0 \u00a0 \u00a0 \u00a0 ……. (5)
        \nat t = 0,\u00a0 x = 1 cm, \u03c9 = \u03c0 s-1<\/sup>
        \nSubstituting in (5) we get
        \n1 = B sin(\u03b1)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0…… (6)
        \nDifferentiate (5) w.r.t. \u2018 t\u2019
        \n\u21d2 v = \\(\\frac{\\mathrm{d} \\mathrm{x}}{\\mathrm{dt}}\\)= B \u03c9 cos (\u03c9 t + \u03b1) dt
        \nat t = 0, v = \u03c9 = \u03c0
        \n\u21d2 1 = B cos(\u03b1)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0…… (7)
        \nDivide(7) by (6)
        \nwe get tan \u03a6 = 1
        \n\u21d2 \u03a6 = \u03c0\/4 is the initial phase
        \nSubstitute \u03a6 in equation (6) we get
        \n1 = B sin (\u03c0\/4) \u21d2 B = \\(\\sqrt{2} \\mathrm{cm}\\)<\/p>\n

        Question 8.
        \nA spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body?
        \nAnswer:
        \nMaximum length Ymax<\/sub> = 0.2m
        \nMaximum weight Mmax<\/sub> = 50 kg
        \nWe know that,
        \nF = Mg = ky, at M = Mmax<\/sub>, Y = Ymax<\/sub>
        \n\"1st
        \n\u21d2 m = 22.36 kg
        \nWeight of the body = 22.36 \u00d7 9.8 = 219.1 N<\/p>\n

        Question 9.
        \nA spring having a spring constant. 1200 N m-1<\/sup> is mounted on a horizontal table as shown in Fig. A mass of 3 kg is attached to the free end of the spring. The mass Is then pulled sideways to a distance of 2.0 cm and released.
        \n\"1st
        \nDetermine<\/p>\n

          \n
        1. the frequency of oscillations,<\/li>\n
        2. maximum acceleration of the mass, and<\/li>\n
        3. the maximum speed of the mass.<\/li>\n<\/ol>\n

          Answer:
          \nGiven:
          \nK = 1200 Nm-1<\/sup>, m = 3 kg, a = 2 cm
          \n1.\u00a0 Frequency of Oscillation
          \nWe know that T = \\(2 \\pi \\sqrt{\\frac{\\mathrm{m}}{\\mathrm{K}}}=2 \\pi \\sqrt{\\frac{3}{1200}}\\).
          \n\u21d2 T = 0.3144 s
          \n\u21d2 f = \\(\\frac{1}{\\mathrm{T}}\\) = 3.18\/s<\/p>\n

          2. Maximum acceleration of mass (\u03b1max<\/sub>) maximum acceleration is obtained when y = a
          \n\u21d2 m \u03b1max<\/sub> = Ka
          \n\u21d2 \u03b1max<\/sub> = \\(\\frac{\\mathrm{Ka}}{\\mathrm{m}}=\\frac{1200 \\times 0.02}{3}\\) = 8 ms-2
          \n<\/sup>
          \n3. Maximum speed of the mass
          \nVmax<\/sub> = a \u03c9 = \\(a \\sqrt{\\frac{k}{m}}=0.02 \\times \\sqrt{\\frac{1200}{3}}\\)
          \nVmax<\/sub> = 0.4 ms-1<\/sup><\/p>\n

          Question 10.
          \nIn Exercise 14.9, let us take the position of mass when the spring is unstretched as x=0, and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t= 0), the mass is
          \n1. at the mean position,
          \n2. at the maximum stretched position, and
          \n3. at the maximum compressed position.
          \nIn what way do these functions for SHM differ from each other, in frequency, in amplitude or the Initial phase?
          \nAnswer:
          \nGiven a = 2 cm k = 1200Nm-1<\/sup> m = 3 kg
          \n\u21d2 \\(\\omega=\\sqrt{\\frac{\\mathrm{k}}{\\mathrm{m}}}=\\sqrt{\\frac{1200}{3}}=20 \\mathrm{s}^{-1}\\)
          \nlet equation of S H M \u21d2 x = A sin \u03c9 t
          \n1. time is measured from mean position.
          \nx = A sin \u03c9 t
          \n\u21d2 x = 2 sin 20 t<\/p>\n

          2.\u00a0 at maximum structured position phase angle is \u03c0\/2
          \n\u21d2 x = a sin (\u03c9 t + \u03c0\/2)
          \n\u21d2 x = 2 cos (20t)<\/p>\n

          3. At maximum compressed position phase angle is 3 \u03c0\/2
          \nx= a sin (\u03c9 t + 3 \u03c0\/2)
          \n\u21d2 x = – a cos \u03c9 t<\/p>\n

          \"KSEEB<\/p>\n

          Question 11.
          \nThe figure below correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e. clockwise or anticlockwise) are indicated on each figure.
          \n\"1st
          \nObtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case.
          \nAnswer:
          \n\"1st
          \n(a) Let A be any point on the circle shown in fig (a). Draw AM \u22a5 to x-axis. Point M refers to x – projection of the radius vector.
          \nNow, \u2220POA = \u03b8 = \u03c9t, T = 2s OA = 3cm \u21d2 \u2220OAM =\u03b8=\u03c9t (\u2235 Alternate angles)
          \n\"1st
          \n\u21d2 x = 3 sin \\(\\frac{2 \\pi}{\\mathrm{T}}\\)t(cm)
          \n\u21d2 x = – 3 sin \u03c0 t (cm)<\/p>\n

          (b) let A be any point on the circles of fig. (b). From A draw AM \u22a5 to x-axis
          \nNow \u2220MOA= \u03b8 = \u03c9t
          \n\"1st<\/p>\n

          Question 12.
          \nPlot the corresponding reference circle for each of the following simple harmonic motions. Indicate the Initial (t =0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anti-clockwise In every case: (x Is In cm and t is in s).<\/p>\n

            \n
          1. x = – 2 sin (3t + \u03c0\/3)<\/li>\n
          2. x = cos(\u03c0\/6 – t)<\/li>\n
          3. x = 3 sin (2 \u03c0 t + \u03c0\/4)<\/li>\n
          4. x = 2 cos \u03c0 t<\/li>\n<\/ol>\n

            Answer:
            \n1.
            \n\"1st
            \n\"1st
            \n2.\u00a0 x = cos(\u03c0\/6 – t)
            \nradius r = 1 cm
            \nat t = 0, x = cos (\u03c0\/6) = \\(\\frac{\\sqrt{3}}{2} \\mathrm{cm}\\),
            \n\u03a6 = – \u03c0\/6 = – 30\u00b0
            \n\u03c9 = 1 rad \/ s
            \n\"1st
            \n3.
            \n\"1st
            \n4. x = 2 cos \u03c0 t
            \nr = 2 cm
            \n\u03c9 = \u03c0 rad \/ s
            \nat t = 0 , x = 2 cm
            \n\"1st<\/p>\n

            Question 13.
            \nFigure (a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Fig. (b) is stretched by the same force F.
            \n\"1st
            \n\"1st<\/p>\n

              \n
            1. What is the maximum extension of the spring in the two cases?<\/li>\n
            2. If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case?<\/li>\n<\/ol>\n

              Answer:
              \n1. Consider figure (a)
              \nlet y be the extension produced in the spring F = ky
              \nconsider fig (b) each mass acts as if it is fixed w.r.t the other
              \n\u21d2 F = ky\u00a0 \u21d2 y = F\/k<\/p>\n

              2. consider fig (a)
              \nF = – ky
              \nma = – ky\u00a0 \u00a0a = \\(\\frac{-\\mathrm{k}}{\\mathrm{m}} \\mathrm{y}\\)
              \n\u21d2 \u03c92<\/sup> = \\(\\frac{\\mathrm{k}}{\\mathrm{m}}\\)
              \nTherefore, \\(\\mathrm{T}=\\frac{2 \\pi}{\\omega}=2 \\pi \\sqrt{\\frac{\\mathrm{m}}{\\mathrm{k}}}\\)
              \nConsider fig (b)
              \nlet us assume (1) as centre of system and 2 springs each of length 1\/2 attached to two masses. So k’ is the spring factor of each spring.
              \nk’ = 2k
              \n\"1st
              \n\u21d2 \\(\\mathrm{T}=\\frac{2 \\pi}{\\omega}\\)
              \n\\(\\mathrm{T}=2 \\pi \\sqrt{\\frac{\\mathrm{m}}{2 \\mathrm{k}}}\\) is the period of oscillation in the case of (b).<\/p>\n

              Question 14.
              \nThe piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad\/ min, what is its maximum speed?
              \nAnswer:
              \nStroke = 1 m
              \nAmplitude = \\(\\frac{\\text { stroke }}{2}\\) = 1\/2 m
              \n\u03c9 = 200 rad \/ min
              \nVmax<\/sub> = \u03c9A
              \n= 200 \u00d7 1\/2
              \n= 100 m \/ min
              \nVmax<\/sub> = 1.67 m\/s<\/p>\n

              Question 15.
              \nThe acceleration due to gravity on the surface of moon is 1.7 m s-2<\/sup>. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s? (g on the surface of earth is 9.8 m s-2<\/sup>)
              \nAnswer:
              \ngm<\/sub> =1.7 ms-2\u00a0 <\/sup>ge<\/sub> = 9.8 ms-2\u00a0 <\/sup>Te<\/sub> = 3.5 s
              \n\"1st<\/p>\n

              Question 16.
              \nAnswer the following questions :<\/p>\n

                \n
              1. Time period of a particle in SHM depends on the force constant k and mass m of the particle: T =\\(2 \\pi \\sqrt{\\frac{m}{k}}\\) . A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?<\/li>\n
              2. The motion of a simple pendulum is approximately simple harmonic for small-angle oscillations. For larger angles of oscillation, a more involved analysis shows that T is greater than \\(2 \\pi \\sqrt{\\frac{1}{g}}\\) . Think of a qualitative argument to appreciate this result.<\/li>\n
              3. A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall?<\/li>\n
              4. What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity?<\/li>\n<\/ol>\n

                Answer:<\/p>\n

                1. In case of a simple pendulum, k is directly proportional to m. Hence ratio of m \/ k is a constant. Hence time period doesn\u2019t depend on mass.<\/p>\n

                2. Restoring force that brings body of pendulum back to its mean position F = – mg sin \u03b8
                \nForm small \u03b8, sin \u03b8 \u2248 \u03b8 = \\(\\frac{\\mathrm{y}}{1}\\)
                \n\"1st
                \nIf approximation for sin \u03b8 = \u03b8 is not taken into account then Time period T > \\(2 \\pi \\sqrt{\\frac{1}{\\mathrm{g}}}\\) This happens when \u03b8 is not small.<\/p>\n

                3. Wristwatch work on the principle of spring action. Hence acceleration due to gravity plays no role in the functioning of the wristwatch hence it gives correct time during free fall.<\/p>\n

                4. During free-fall acceleration, due to gravity is zero hence the pendulum will not vibrate (i.e.: frequency of oscillation is zero).<\/p>\n

                \"KSEEB<\/p>\n

                Question 17.
                \nA simple pendulum of length I and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations In a radial direction about its equilibrium position, what will be its time period?
                \nAnswer:
                \nBody of the pendulum is under the action of two accelerations, acceleration due to gravity \u2018g\u2019 and centripetal acceleration \\(\\alpha=\\frac{v^{2}}{R}\\)
                \neffective acceleration = \\(\\alpha^{\\prime}=\\sqrt{\\alpha^{2}+g^{2}}\\)
                \n\"1st<\/p>\n

                Question 18.
                \nA cylindrical piece of cork of density of base area A and height h floats in a liquid of density \\(\\rho_{\\mathrm{t}}\\). The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a \\(\\mathrm{T}=2 \\pi \\sqrt{\\frac{\\mathrm{h} \\rho}{\\rho_{1} \\mathrm{g}}}\\) where \u03c1 is the density of cork. (Ignore damping due to viscosity of the liquid).
                \nAnswer:
                \nInitially in equilibrium,
                \nweight of cork = weight of water displaced
                \n\"1st
                \nWhen cork is pushed in, then the restoring force acting on it is :
                \nf = – weight of the portion dipped after pushing
                \n\"1st
                \n\"1st<\/p>\n

                Question 19.
                \nOne end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic, motion.
                \nAnswer:
                \nRestoring force acting on the liquid when suction pump is removed is,
                \nf = – mg\u00a0 \u21d2 f= -(A \u00d7 2y)\u03c1 \u00d7 g
                \nwhere, A = Cross – section area of U tube
                \n\u03c1 = density of liquid
                \n\u21d2 f = – 2A\u03c1gy
                \nacceleration produced in liquid column
                \na = f\/mass of liquid
                \n\"1st
                \nhence it is a SHM.<\/p>\n

                Question 20.
                \nAn air chamber of volume V has a neck area of cross-section into which a ball of mass m just fits and can move up and down without any friction. Show that when the ball is pressed down a little and released, it executes SHM. Obtain an expression for the time period of oscillations assuming pressure-volume variations of air to be isothermal.
                \n\"1st
                \nAnswer:
                \nVolume = V
                \nmass of ball = m
                \ncross-sectional Area = A
                \ninitial pressure on either side of the ball = atmospheric pressure = P
                \nLet the charge in volume of air when ball is pressed be \u2206V
                \n\u2206V = Ay (y = displacement)
                \nNow, Bulk modulus of elasticity
                \n\"1st
                \n\"1st
                \nThe above equation is of the form,
                \n\"1st
                \nhence it is in SHM.<\/p>\n

                Question 21.
                \nYou are riding in an automobile of mass 3000 kg. Assuming that you are examining the oscillation characteristics of Its suspension system. The suspension sags 15 cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50% during one complete oscillation. Estimate the values of<\/p>\n

                  \n
                1. the spring constant k and<\/li>\n
                2. the damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports 750 kg.<\/li>\n<\/ol>\n

                  Answer:
                  \n1. Total mass = 3000 kg
                  \nmass supported by each wheel = 750 kg
                  \ny = 0.15 m
                  \nWe know that,
                  \nmg = kg
                  \n\"1st
                  \n2.
                  \n\"1st<\/p>\n

                  Question 22.
                  \nShow that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the
                  \nsame period.
                  \nAnswer:
                  \nLet the particle executing SHM starts from mean position.
                  \nDisplacement can be given by
                  \nx = Asin \u03c9 t
                  \nVelocity v = A \u03c9 cos\u03c9t
                  \nKinetic energy over one complete cycle
                  \n\"1st
                  \n\"1st
                  \n\"1st<\/p>\n

                  Question 23.
                  \nA circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillators is found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant \u03b1 is defined by the relation J= – \u03b1 \u03b8, where j is the restoring couple and \u03b8 the angle of twist).
                  \nAnswer:
                  \nWe know that,
                  \n\"1st<\/p>\n

                  Question 24.
                  \nA body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is<\/p>\n

                    \n
                  1. 5 cm<\/li>\n
                  2. 3 cm<\/li>\n
                  3. 0 cm.<\/li>\n<\/ol>\n

                    Answer:
                    \nGiven,
                    \nr = 5 cm
                    \nT = 0.2 s
                    \n\\(\\omega=\\frac{2 \\pi}{T}=\\frac{2 \\pi}{0.2}=10 \\pi \\mathrm{rad} \/ \\mathrm{s}\\)
                    \nacceleration A = – \u03c92<\/sup>y
                    \nvelocity V = \\(\\omega \\sqrt{r^{2}-y^{2}}\\)
                    \n1. y = 5 cm
                    \nA = – (10\u03c0)2<\/sup> \u00d7 0.05
                    \n\u21d2 A = 0.493 m \/ s2<\/sup>
                    \nV = \\(10 \\pi \\sqrt{(0.05)^{2}-(0.05)^{2}}=0\\)<\/p>\n

                    2. y = 3 cm
                    \nA = (10\u03c0)2<\/sup> \u00d7 0.03
                    \n\u21d2 A = – 0.296 m \/ s2<\/sup>
                    \nV= \\(10 \\pi \\sqrt{(0.05)^{2}-(0.03)^{2}}\\)
                    \nV = 0.4 \u03c0m\/s<\/p>\n

                    3. y = 0 cm
                    \nA = 0
                    \nV= \\(10 \\pi \\sqrt{(0.05)^{2}-0}\\)
                    \n= 0.5 \u03c0m\/s<\/p>\n

                    \"KSEEB<\/p>\n

                    Question 25.
                    \nA mass attached to a spring is freeto oscillate, with angular velocity \u03c9, in a horizontal plane without friction or damping. It is pulled to a distance x0<\/sub> and pushed towards the centre with a velocity v0<\/sub> at time t = 0. Determine the amplitude of the resulting oscillations in terms of the parameters \u03c9, x0<\/sub> and v0<\/sub>.
                    \n[Hint : Start with the equation x = a cos (\u03c9t+\u03b8) and note that the initial velocity is negative.]
                    \nAnswer:
                    \nLet x = a cos (\u03c9t + \u03b8)
                    \n\u2234 v = \\(\\frac{d x}{d t}\\) = – a \u03c9 sin (\u03c9t + \u03b8)
                    \nWhen t = 0, x = x0<\/sub> and \\(\\frac{d x}{d t}\\) = – v0<\/sub>
                    \n\u21d2 xo<\/sub> = a cos \u03b8 ….. (1) and
                    \n– v0<\/sub> = – a \u03c9 sin \u03b8
                    \n\u21d2 a sin \u03b8 = \\(\\frac{\\mathrm{v}_{0}}{\\omega}\\) …..(2)
                    \nSquarring and adding (1) and (2),
                    \na2<\/sup> = \\(=x_{0}^{2}+\\frac{v_{0}^{2}}{\\omega^{2}} \\Rightarrow a=\\sqrt{x_{0}^{2}+\\frac{v_{0}^{2}}{\\omega^{2}}}\\)<\/p>\n

                    1st PUC Physics Oscillations One Mark Questions and Answers<\/h3>\n

                    Question 1.
                    \nMention the relation between period and frequency of periodic motion.
                    \nAnswer:
                    \nFrequency f = \\(\\frac{1}{T}\\)<\/p>\n

                    Question 2.
                    \nA particle executes simple harmonic motion. At what point on its path is the acceleration maximum?
                    \nAnswer:
                    \nAcceleration is maximum at a point of maximum displacement from the mean position.<\/p>\n

                    Question 3.
                    \nAt what position the KE of an oscillating simple pendulum Is maximum?
                    \nAnswer:
                    \nKinetic energy is maximum at the mean position.<\/p>\n

                    Question 4.
                    \nWhat is the condition for motion of a particle to be SHM?
                    \nAnswer:
                    \nAcceleration should be proportional to displacement and always directed towards mean position.<\/p>\n

                    Question 5.
                    \nWhat is Oscillation?
                    \nAnswer:
                    \nTo and for motion in the same path is called Oscillation.<\/p>\n

                    Question 6.
                    \nHow will the period of a simple pendulum change when its length is doubled?
                    \nAnswer:
                    \nTnew<\/sub> = \\(\\sqrt{2}\\) Told<\/sub><\/p>\n

                    \"KSEEB<\/p>\n

                    Question 7.
                    \nWill a pendulum\u2019s time period increases or decreases when taken to the top of the mountain?
                    \nAnswer:
                    \nIncreases, g decreases as high altitude and \\(\\mathrm{T} \\alpha \\frac{1}{\\sqrt{\\mathrm{g}}}\\)<\/p>\n

                    Question 8.
                    \nTwo simple pendulum of equal length cross each other at mean position. What is their phase difference?
                    \nAnswer:
                    \n180\u00b0 (\u03c0 radians)<\/p>\n

                    Question 9.
                    \nHow many times in one vibration, KE and PE become maximum?
                    \nAnswer:
                    \nTwo.<\/p>\n

                    Question 10.
                    \nWhen is the tension maximum in the spring of a simple pendulum?
                    \nAnswer:
                    \nMean position.<\/p>\n

                    Question 11.
                    \nA spring of spring constant k Is cut Into two equal parts. What is the spring constant of each part?
                    \nAnswer:
                    \n2 K<\/p>\n

                    Question 12.
                    \nWhat Is the phase difference between the displacement and velocity in a SHM?
                    \nAnswer:
                    \n90\u00b0 (\u03c0\/2 radians)<\/p>\n

                    Question 13.
                    \nState force law for a SHM.
                    \nAnswer:
                    \nForce, F = m\u03c92<\/sup>x
                    \nm = mass
                    \n\u03c9 = angular speed
                    \nx = displacement<\/p>\n

                    Question 14.
                    \nA pendulum is making one oscilla\u00action in every two seconds. What is the frequency of oscillation?
                    \nAnswer:
                    \nf = \\(\\frac{1}{T}=\\frac{1}{2} s^{-1}\\)<\/p>\n

                    \"KSEEB<\/p>\n

                    Question 15.
                    \nWhat is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity?
                    \nAnswer:
                    \nFrequency is 0 because the acceleration zero during free fall.<\/p>\n

                    Question 16.
                    \nA simple pendulum Is inside a spacecraft. What should be its time period vibration?
                    \nAnswer:
                    \nPendulum does not oscillate.<\/p>\n

                    Question 17.
                    \nWhat are isochronous vibrations?
                    \nAnswer:
                    \nWhen the time period is independent of amplitude such an oscillation is called isochronous.<\/p>\n

                    1st PUC Physics Oscillations Two Marks Questions and Answers<\/h3>\n

                    Question 1.
                    \nDefine simple harmonic motion (SHM). Give an example.
                    \nAnswer:
                    \nA particle is said to have SHM if the acceleration of the particle is directly proportional to its displacement from the mean position and directed towards the mean position.
                    \nE.g.:<\/p>\n

                      \n
                    1. Vertical oscillations of a loaded spring.<\/li>\n
                    2. Oscillations of bob of a simple pendulum.<\/li>\n
                    3. Vibrations of string of musical instruments.<\/li>\n
                    4. Motion of air particles during the propagation of sound waves.<\/li>\n
                    5. Vibration of a tuning fork.<\/li>\n<\/ol>\n

                      Question 2.
                      \nMention expression for velocity and acceleration of a particle executing SHM.
                      \nAnswer:
                      \nVelocity of a particle executing SHM is v = \\(\\omega \\sqrt{A^{2}-y^{2}}\\)
                      \nAcceleration of a particle executing SHM is a = – \u03c92<\/sup>y
                      \ny is the displacement of a particle from its mean position in t seconds. A is the amplitude and \u03c9 is the angular frequency.<\/p>\n

                      Question 3.
                      \nMention expression for K.E, P.E and total energy of a particle executing SHM.
                      \nAnswer:
                      \nPotential energy of particle at any instant t second is Ep<\/sub> = 1\/2 m \u03c92<\/sup> y2<\/sup>.
                      \nKinetic energy of particle at any instant t second is, EK<\/sub> = 1\/2 m \u03c92<\/sup> (A2<\/sup> – y2<\/sup>).
                      \nTotal energy of particle at any instant is constant E = EK<\/sub> + Ep<\/sub> = 1\/2 m \u03c92<\/sup> A2<\/sup>.
                      \nA is the amplitude and w is the angular frequency and m is mass of the particle y is the displacement of a particle from its mean position in t seconds.<\/p>\n

                      \"KSEEB<\/p>\n

                      Question 4.
                      \nThe acceleration of a particle executing SHM Is 20ms-1<\/sup> at a distance of 5 m from the mean position. Calculate its time period and frequency.
                      \nAnswer:
                      \na = 20m\/s2<\/sup>, y = 5m.
                      \nWe know that a = | \u03c92<\/sup> y |
                      \n\u21d2 \\(\\omega=\\sqrt{\\frac{a}{y}}=\\sqrt{\\frac{20}{5}}\\)= 2 rad\/s or 2\u03c0f = 2
                      \nFrequency f = \\(\\frac{2}{2 \\times 3.14}\\) = 0.3185 Hz
                      \nPeriod T = \\(\\frac{1}{f}=\\frac{1}{0.3185}\\) = 3.14s<\/p>\n

                      Question 5.
                      \nDifferentiate between forced oscillations and resonance.
                      \nAnswer:
                      \n1. Forced Oscillations:
                      \nA body oscillates with the help of external periodic force with a frequency different from natural frequency of body.<\/p>\n

                      2. Resonance:
                      \nA body oscillating with its natural frequency with the help of external periodic force whose frequency is equal to natural frequency of body.<\/p>\n

                      Question 6.
                      \nTwo linear, simple harmonic motion of equal amplitudes and frequencies \u03c9 and 2\u03c9 are impressed on a particle along the axis of X and Y respectively. If the initial phase difference is \u03c0\/2, then find the resultant path followed by particle
                      \nAnswer:
                      \nLet
                      \nX = Asin\u03c9t …… (1)
                      \nY= Acos2\u03c9t …….(2)
                      \n(1) and (2) represent SHM with equal amplitude and phase difference \u03c0\/2 with frequencies \u03c9 and 2\u03c9.<\/p>\n

                      Question 7.
                      \nDerive an expression for K.E and P.E of a particle SHM.
                      \nAnswer:
                      \nWe know that, Potential energy
                      \nU = 1\/2 kx2<\/sup>
                      \nFor a particle in SHM, k = m\u03c92<\/sup> and x = Asin \u03c9t
                      \n\u21d2 U = 1\/2 m\u03c92<\/sup> A2<\/sup> sin2<\/sup> \u03c9t
                      \nKinetic energy, k= 1\/2 mv2<\/sup>
                      \nnow, v = \u03c9y = \u03c9A cos \u03c9t
                      \n\u21d2 K= 1\/2 m\u03c92<\/sup> A2<\/sup>cos2<\/sup>\u03c9t<\/p>\n

                      Question 8.
                      \nThe amplitude of a oscillating simple pendulum Is doubled. What will be its effect on<\/p>\n

                        \n
                      1. periodic time<\/li>\n
                      2. total energy<\/li>\n
                      3. maximum velocity<\/li>\n<\/ol>\n

                        Answer:<\/p>\n

                          \n
                        1. Periodic time does not change. T is independent of amplitude.<\/li>\n
                        2. Total energy, T.E = 1\/2 m\u03c92<\/sup> A2<\/sup>
                          \nA is doubled \u21d2 T.Enew<\/sub> = 4 TEold<\/sub><\/li>\n
                        3. Maximum velocity, Vmax<\/sub> = \u03c9A
                          \nA is doubled \u21d2 Vmax<\/sub> is doubled.<\/li>\n<\/ol>\n

                          Question 9.
                          \nThe frequency of oscillations of a mass m suspended by a spring is V1<\/sub>. If the length of spring is cut to one half, the same mass oscillates with frequency V2<\/sub>. Calculate V2<\/sub>\/V1<\/sub>.
                          \nAnswer:
                          \nV1<\/sub> = \\(\\frac{1}{2 \\pi} \\sqrt{\\frac{k}{m}}\\)
                          \nwhere K is the spring constant and m is the mass
                          \nV2<\/sub> = \\(\\frac{1}{2 \\pi} \\sqrt{\\frac{2 \\mathrm{k}}{\\mathrm{m}}}\\)
                          \n\u2234 \\(\\frac{V_{2}}{V_{1}}=\\sqrt{2}\\)<\/p>\n

                          \"KSEEB<\/p>\n

                          Question 10.
                          \nDiscuss some important characteristics of wave motion.
                          \nAnswer:<\/p>\n

                            \n
                          1. Wave transports energy.<\/li>\n
                          2. particles are not transported.<\/li>\n
                          3. Elasticity and inertia determine the motion of particle distribution.<\/li>\n<\/ol>\n

                            Question 11.
                            \nDifferentiate between free oscillations and forced oscillations with the help of examples.
                            \nAnswer:
                            \nConsider a pendulum in free space (without air) it oscillates freely. This is free oscillation. Consider another pendulum in a viscous liquid it oscillates only if there is external force this is forced oscillation.<\/p>\n

                            Question 12.
                            \nList any two characteristics of SHM.
                            \nAnswer:<\/p>\n

                              \n
                            1. SHM is always directed towards mean position.<\/li>\n
                            2. Acceleration is directly proportioned to displacement but opposite in direction.<\/li>\n<\/ol>\n

                              1st PUC Physics Oscillations Three Marks Questions and Answers<\/h3>\n

                              Question 1.
                              \nDefine the terms<\/p>\n

                                \n
                              1. amplitude<\/li>\n
                              2. period<\/li>\n
                              3. frequency<\/li>\n
                              4. phase related with a particle executes SHM.<\/li>\n<\/ol>\n

                                Answer:
                                \n1. Amplitude:
                                \nMaximum displacement of the particle from the mean position is called amplitude.<\/p>\n

                                2. Period :
                                \nTime taken by the particle to complete one oscillation is called time period. (T)<\/p>\n

                                3. Frequency :
                                \nThe number of oscillations completed by the particle in one second is called frequency (f)<\/p>\n

                                4. Phase:
                                \nPhase of a particle is defined as the fraction of the time period that has elapsed since the particle last passed through its mean position in the positive direction.<\/p>\n

                                \"KSEEB<\/p>\n

                                Question 2.
                                \nFor an oscillating pendulum, establish the relation \\(\\frac{d^{2} \\theta}{d t^{2}}=-\\omega^{2} \\theta\\) where \\(\\omega=\\sqrt{\\frac{g}{1}} \\theta\\) = small angular displacement.
                                \nAnswer:
                                \n\"1st
                                \nrestoring force Fr<\/sub> = – mg sin \u03b8
                                \ntorque on pendulum \\(\\tau=\\mathrm{I} \\alpha=\\mathrm{ml}^{2} \\alpha\\)
                                \nrestoring torque = torque on pendulum
                                \n\"1st
                                \nWe know that = \\(\\frac{\\mathrm{d}^{2} \\theta}{\\mathrm{dt}^{2}}=\\alpha\\) and \\(\\omega=\\sqrt{\\frac{g}{1}}\\)
                                \n\u21d2 \\(\\frac{\\mathrm{d}^{2} \\theta}{\\mathrm{dt}^{2}}=-\\omega^{2} \\theta\\)<\/p>\n

                                Question 3.
                                \nA body of mass 1 kg is suspended from a weightless spring having force constant 600 Nm-1<\/sup>. Another body of mass 0.5 kg moves vertically upwards lift the suspended body with a velocity of 3 ms-1<\/sup> and tets embedded in it. Find the frequency of oscillation and amplitude of motion.
                                \nAnswer:
                                \nTotal mass = (1 + 0.5)kg = 1.
                                \nk = 600 nm-1<\/sup>
                                \nfrequency of oscillations
                                \n\"1st
                                \nLet V1<\/sub> be velocity of mass after collision
                                \n\u21d2 m1<\/sub> V2<\/sub> = (m1<\/sub> + m2,<\/sub>) V1<\/sub>
                                \n\u21d2 0.5 \u00d7 3 = 1.5 V1<\/sub>
                                \n\u21d2 V1<\/sub> = 1 m\/s
                                \nAccording to law of conservation of energy
                                \n(PE)max<\/sub> = (KE)max<\/sub>
                                \n\\(\\frac{1}{2} m v_{1}^{2}=1\/2^{K A^{2}}\\)
                                \n\\(\\frac{1}{2} 1.5 \\times 1^{2}=\\frac{1}{2} 600 \\mathrm{A}^{2}\\)
                                \n\u21d2 A = 5 cm<\/p>\n

                                Question 4.
                                \nThe bottom of a dip on a road has a radius of curvature R. A rickshaw of mass M left a little away from the bottom oscillates about this dip. Deduce an expression for the period of oscillation.
                                \nAnswer:
                                \n\"1st
                                \nLet the rikshaw of mass M be at any point P in the dip of radius R. Let O be the centre of this circular path. This case is similar to that of a simple pendulum and assume that e is small.
                                \nrestoring force
                                \nF = -mg sin\u03b8
                                \nF = – mg \u03b8
                                \ndisplacement of rickshaw = R \u03b8
                                \n\"1st<\/p>\n

                                Question 5.
                                \nFind an expression for body vibrating in SHM.
                                \nAnswer:
                                \nLet the SHM equation be
                                \nx = Asin \u03c9 t
                                \n\"1st
                                \nPotential energy is given by
                                \n= 1\/2 kx2<\/sup> = 1\/2 m \u03c9 2<\/sup> k2<\/sup>
                                \n= 1\/2 m\u03c92<\/sup> A2<\/sup> sin2<\/sup> \u03c9 t
                                \nKinetic energy = 1\/2 mv2<\/sup>
                                \n= 1\/2 m \u03c9 2<\/sup> A2<\/sup> cos2<\/sup> \u03c9 t
                                \nTotal energy\u00a0 \u00a0 = PE + KE
                                \n= 1\/2 m\u03c92<\/sup> A2<\/sup> sin2<\/sup> \u03c9 t + 1\/2 m\u03c92<\/sup> A2<\/sup> cos2<\/sup> \u03c9 t
                                \n= 1\/2 m\u03c9 2<\/sup> A2<\/sup><\/p>\n

                                Question 6.
                                \nShow that when a particle is moving in SHM its velocity at a distance \\(\\frac{\\sqrt{3}}{2}\\) its amplitude from the central position is half its velocity in central position.
                                \nAnswer:
                                \nFor a particle in SHM
                                \n\\(\\mathrm{V}=\\omega \\sqrt{\\mathrm{A}^{2}-\\mathrm{x}^{2}}\\)
                                \n\"1st<\/p>\n

                                Question 7.
                                \nA body oscillates with SHM according to the equation.
                                \nx(t) = 5cos (2\u03c0 t + \u03c0 \/4) Where x is in meters and t is in seconds calculate,<\/p>\n

                                  \n
                                1. displacement att = 0<\/li>\n
                                2. Angular frequency<\/li>\n
                                3. Vmax<\/sub><\/li>\n<\/ol>\n

                                  Answer:
                                  \n1. displacement at t = 0
                                  \nx(0) = 5cos(2\u03c0 0 + \u03c0\/4)
                                  \n\u21d2 x = \\(\\frac{5}{\\sqrt{2}} m\\)<\/p>\n

                                  2. Angular frequency
                                  \n\u03c9 = 2\u03c0 radian\/s<\/p>\n

                                  3. Vmax = a\u03c9 = 2\u03c0 \u00d7 5 = 10\u03c0 m\/s<\/p>\n

                                  \"KSEEB<\/p>\n

                                  Question 8.
                                  \nWhat is spring factor? Find its value in case of two springs connected in<\/p>\n

                                    \n
                                  1. series<\/li>\n
                                  2. Parallel<\/li>\n<\/ol>\n

                                    Answer:
                                    \nSpring factor (k):
                                    \nForce acting for unit extension produced is called spring factor.
                                    \n1. When two springs connected in series,
                                    \n\"1st
                                    \nforce experienced by both springs is same.
                                    \nTotal extension (x) is sum of individual extension.
                                    \n\u21d2 x = x1<\/sub> + x2<\/sub>
                                    \n\\(\\frac{F}{-K_{e q}}=\\frac{F}{-K_{1}}+\\frac{F}{-K_{2}}\\) \u21d2 1\/Keq<\/sub> = 1\/K1<\/sub> + 1\/K2<\/sub>
                                    \nKeq<\/sub> = \\(\\frac{\\mathrm{K}_{1} \\mathrm{K}_{2}}{\\mathrm{K}_{1}+\\mathrm{K}_{2}}\\)<\/p>\n

                                    2. When two springs are in parallel,
                                    \n\"1st
                                    \nNet force experienced is the sum of force experienced in both springs. But the extension will be the same.
                                    \nFeq<\/sub> = F1<\/sub> + F2<\/sub>
                                    \nKeq<\/sub>x =K1<\/sub> X + K2<\/sub>X
                                    \n\u21d2 Keq<\/sub> = K1<\/sub> + K2<\/sub><\/p>\n

                                    Question 9.
                                    \nExplain the relation in phase between displacement, velocity, and acceleration in SHM, graphically as well as theoretically.
                                    \nAnswer:
                                    \nLet displacement x = Asin\u03c9t
                                    \nVelocity v= \u03c9 Acos\u03c9 t
                                    \nacceleration a = – \u03c92<\/sup> Asin \u03c9 t = \u03c92<\/sup> Asin(\u03c9 t + \u03c0 )
                                    \nDisplacement and velocity have phase difference of \u03c0\/2 radians.
                                    \nv and \u03b1 has \\(\\frac{\\pi}{2}\\) radians of phase difference. \u03b1 ans x has \u03c0 radians of phase difference.
                                    \nAssume \u03c9 >1
                                    \n\"1st<\/p>\n

                                    Question 10.
                                    \nFor a particle in SHM, the displacement x of the particle at a function of time t is given as x = Asin(2\u03c0t) …..(1) where x is in centimeters and t is in seconds. Let the time taken by the particle to travel from x=0 to x = A\/2 beT1<\/sub> and the time taken to travel from x = A\/2 to x = A be T2<\/sub>. Find T1<\/sub>\/T2<\/sub>
                                    \nAnswer:
                                    \n\u03c9 =2\u03c0 = \\(\\frac{2 \\pi}{\\mathrm{T}}\\) \u21d2 T= 1 s
                                    \nat t = 0 , x = 0
                                    \nnow if x = A\/2
                                    \n\"1st
                                    \nnow in a SHM time taken from 0 to A is \\(\\frac{T}{4}\\)
                                    \n\u21d2 time taken for x = A\/2 to A is \\(\\frac{T}{4}\\) -T\/12
                                    \n\"1st<\/p>\n

                                    Question 11.
                                    \nWhat is SHM? Show that in SHM acceleration is directly proportional to Its displacement at a given instant.
                                    \nAnswer:
                                    \nSimple Harmonic Motion is a type of motion in which displacement is always directed towards mean position and acceleration is directly proportional to displacement and opposite in direction.
                                    \nLet a SHM be represented by
                                    \nx = Asin \u03c9t
                                    \n\u21d2 \\(\\frac{\\mathrm{d} x}{\\mathrm{dt}}\\) = A \u03c9 cos \u03c9t
                                    \n\"1st
                                    \nacceleration is proportional to displacement and opposite in direction.<\/p>\n

                                    Question 12.
                                    \nA 0.2 kg of mass hangs at the end of a spring. When 0.02 kg more mass is added to the end of the spring, it stretched 7cm more. If the 0.02 kg mass is removed, what will be the period of vibration of the system?
                                    \nAnswer:
                                    \nMass added m = 0.02
                                    \nLength stretched x= 7 cm = 0.07 m
                                    \n\"1st
                                    \nTime period T = \\(2 \\pi \\sqrt{\\frac{\\mathrm{m}}{\\mathrm{k}}}\\)
                                    \n\"1st<\/p>\n

                                    Question 13.
                                    \nA particle executes SHM of time period 10 sec. The displacement of particle at any instant is x = 10 sin\u03c9t(cm) Find<\/p>\n

                                      \n
                                    1. Velocity of body 2 s after it passes through mean position.<\/li>\n
                                    2. Acceleration of body 2 s after it passes mean position.<\/li>\n<\/ol>\n

                                      Answer:
                                      \nGiven x = 10 sin \u03c9t T = 10 sec
                                      \nA = 10 cm\u00a0 \u00a0\u03c9 = \\(\\frac{2 \\pi}{10}\\) rad\/s
                                      \nLet at t = 0 body be at the mean position.
                                      \nNow at t = 2
                                      \n1.\u00a0 V = A\u03c9 cos \u03c9t
                                      \n= 10\u03c9 cos(\u03c92)cm\/s
                                      \n\u21d2 \\( V=10 \\times \\frac{2 \\pi}{10} \\cos \\left(\\frac{2 \\pi}{10} 2\\right)\\)
                                      \n\u21d2 V = 1.94 cm\/s<\/p>\n

                                      2. Acceleration
                                      \na = – A\u03c92<\/sup> sin\u03c9t
                                      \n\\(= – 10 \\times \\frac{4 \\pi^{2}}{100} \\sin \\left(\\frac{4 \\pi}{10}\\right)\\)
                                      \n\u21d2 a = – 3.75 cm\/s2<\/sup><\/p>\n

                                      \"KSEEB<\/p>\n

                                      Question 14.
                                      \nWhat is a simple pendulum? Show that the motion of the pendulum is SHM and hence deduce an expression for the time period of pendulum. Also, define Second\u2019s pendulum.
                                      \nAnswer:
                                      \nSimple pendulum is a point mass body suspended by a weightless thread or string from a rigid support about which it is free to oscillate.
                                      \n\"1st
                                      \nNow restoring force acting on body can be given by
                                      \nF = – mgsin\u03b8 (from figure)
                                      \nIf\u00a0 \u03b8 is small
                                      \nF= – mg\u03b8\u00a0 \u00a0 (\u2235 sin \u03b8 \u2248 \u03b8)
                                      \nLet P be any point on the path of pendulum and makes \u2220OMP = \u03b8 which is small and arc OP be x (displacement) then
                                      \n\\(\\dot{\\theta}=\\frac{\\mathrm{OP}}{1}=\\frac{\\mathrm{x}}{1}\\)
                                      \n\u21d2 F = \\(\\frac{-m g x}{1}\\) ……(1)
                                      \n\u21d2 force \u221d displacement and opposite in direction hence an SHM
                                      \nNow this of form F = -kx ……(2)
                                      \n\u21d2 k = mg\/l (from (1) and (2)
                                      \n\"1st<\/p>\n

                                      Question 15.
                                      \nA body of mass ‘m’ suspended from a spring executes SHM. Calculate the ratio of kinetic energy and potential energy of the body when it is at half the amplitude far from the mean position.
                                      \nAnswer:
                                      \nKE of a mass suspend from a spring
                                      \nKE = m\u03c92<\/sup>(a2<\/sup> – y2<\/sup>)
                                      \nPotential energy of a mass suspend from a spring
                                      \nPE = 1\/2 m\u03c92<\/sup> y2<\/sup>
                                      \n\\(\\frac{\\mathrm{KE}}{\\mathrm{PE}}=\\frac{\\left(\\mathrm{a}^{2}-\\mathrm{y}^{2}\\right) 2}{\\mathrm{y}^{2}}\\)
                                      \nat y = a\/2
                                      \n\u21d2 \\(\\frac{\\mathrm{KE}}{\\mathrm{PE}}=\\frac{\\left(\\mathrm{a}^{2}-\\mathrm{a}^{2} \/ 4\\right)^{2}}{\\mathrm{a}^{2} \/ 4}\\)
                                      \n\u21d2 \\(\\frac{\\mathrm{KE}}{\\mathrm{PE}}=3\\)<\/p>\n

                                      Question 16.
                                      \nIf x = a cos \u03c9t + b sin \u03c9t, show that it represents SHM.
                                      \nAnswer:
                                      \nWe have,
                                      \nx = a cos \u03c9t + b sin \u03c9t
                                      \nNow, \\(\\frac{\\mathrm{d} \\mathrm{x}}{\\mathrm{dt}}\\) = – a\u03c9 sin\u03c9t + b\u03c9 cos\u03c9t
                                      \n\\(\\frac{\\mathrm{d}^{2} \\mathrm{x}}{\\mathrm{dt}^{2}}\\) = – a\u03c92<\/sup> cos \u03c9t + b\u03c92<\/sup> sin \u03c9t
                                      \n\u21d2 \\(\\frac{\\mathrm{d}^{2} \\mathrm{x}}{\\mathrm{dt}^{2}}\\) = – \u03c92<\/sup> (a cos \u03c9t+ b sin \u03c9t)
                                      \n\u21d2 \\(\\frac{\\mathrm{d}^{2} \\mathrm{x}}{\\mathrm{dt}^{2}}\\) = – \u03c92<\/sup>x
                                      \n\u21d2 \u03b1 = – \u03c92<\/sup>x
                                      \nHence an SHM<\/p>\n

                                      Question 17.
                                      \nFind the expression for the total energy of a particle executing SHM.
                                      \nAnswer:
                                      \nFor a SHM PE = 1\/2 kx2<\/sup>
                                      \n= 1\/2 m\u03c92<\/sup> x2<\/sup> = 1\/2 m\u03c9 A2<\/sup> sin2<\/sup> \u03c9t
                                      \nKE= 1\/2 mv2<\/sup>
                                      \nv = \\(\\frac{\\mathrm{d} \\mathrm{x}}{\\mathrm{dt}}\\) = – A\u03c9 cos \u03c9t
                                      \n\u21d2 KE = 1\/2 mA2<\/sup>\u03c92<\/sup> cos2<\/sup> \u03c9t
                                      \nTotal energy = KE + PE = 1\/2 m\u03c92<\/sup> A2<\/sup> cos2<\/sup> \u03c9t + 1\/2 m\u03c92<\/sup> sin2<\/sup> \u03c9t
                                      \nTotal energy = 1\/2 m\u03c92<\/sup> A2<\/sup><\/p>\n

                                      1st PUC Physics Oscillations Five Marks Questions and Answers<\/h3>\n

                                      Question 1.<\/p>\n

                                        \n
                                      1. Find the total energy of particle executing SHM?<\/li>\n
                                      2. Show graphically the variation of PE and KE with time In SHM.<\/li>\n
                                      3. What is the frequency of these energies w.r.t the frequency of the particle executing SHM?<\/li>\n<\/ol>\n

                                        Answer:
                                        \n1. PE at any instant
                                        \nPE = 1\/2 kx2<\/sup>
                                        \n= 1\/2 m\u03c92<\/sup> A2<\/sup> sin2<\/sup> \u03c9t
                                        \n(\u2235 k = m\u03c92<\/sup> and x = A sin \u03c9t)
                                        \n\u21d2 PE = 1\/2 m\u03c92<\/sup> A2<\/sup> sin2<\/sup> \u03c9t
                                        \nKE at any instant KE = 1\/2 mv2<\/sup>
                                        \nKE = 1\/2 mA2<\/sup> \u03c92<\/sup> cos2<\/sup> \u03c9t
                                        \nTotal energy = KE + PE
                                        \n\u21d2 Total energy = 1\/2 m\u03c92<\/sup> A2<\/sup>sin2<\/sup> \u03c9t + 1\/2 m\u03c92<\/sup> A2<\/sup> cos2<\/sup> \u03c9t
                                        \n\u21d2 Total energy = 1\/2 m\u03c92<\/sup> A2
                                        \n<\/sup>
                                        \n2. Graph
                                        \n\"1st
                                        \n3. Frequency = \\(\\frac{2}{\\text { period }}=\\frac{2}{\\mathrm{T}}\\)<\/p>\n

                                        1st PUC Physics Oscillations Numerical Problems Questions and Answers<\/strong><\/p>\n

                                        Question 1.
                                        \nA spring compressed by 0.1m develops a restoring force of 10N. A body of mass 4 kg is placed on it. Deduce the<\/p>\n

                                          \n
                                        1. force constant of the spring<\/li>\n
                                        2. depression of the spring under the weight of the body<\/li>\n
                                        3. period of oscillation, if the body is disturbed.<\/li>\n<\/ol>\n

                                          Answer:
                                          \nGiven
                                          \nRestoring force, F = 10N
                                          \nMass of the body, m = 4kg
                                          \ndisplacement, x =0.1m<\/p>\n

                                            \n
                                          1. \\(\\mathrm{k}=\\frac{\\text { force }}{\\text { displacement }}=\\frac{10}{0.1}\\) = 100 Nm-1<\/sup><\/li>\n
                                          2. Depression due to weight \\(=\\frac{\\text { force }}{\\mathbf{k}}=\\frac{4 \\times 10}{100}\\) = 0.4 m<\/li>\n
                                          3. Period of oscillation T = \\(2 \\pi \\sqrt{\\frac{m}{k}}\\) \\(=2 \\pi \\sqrt{\\frac{4}{100}}\\) = 0.4 \u03c0s
                                            \nT = 0.4 \u03c0s<\/li>\n<\/ol>\n

                                            Question 2.
                                            \nA vertical U- tube of uniform crosssection contains water up to a height to 20cm. Calculate the time period of the oscillation of water when it is disturbed.
                                            \nAnswer:
                                            \nThe length of liquid column
                                            \nL = 2 \u00d7 20cm = 40 cm
                                            \nTime period of oscillation
                                            \n\\(=2 \\pi \\sqrt{\\frac{L}{2 g}}=2 \\pi \\sqrt{\\frac{40}{2 \\times 9.80}}\\) = 0.9 s<\/p>\n

                                            \"KSEEB<\/p>\n

                                            Question 3.
                                            \nA cylindrical piece of cork of base area \u2018A’ and height \u2018h\u2019 floats in a liquid of density \\(\\rho_{\\mathrm{e}}\\). The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period of \\(\\mathrm{T}=2 \\pi \\sqrt{\\frac{\\mathrm{h} \\rho}{\\rho_{\\mathrm{e}} \\mathrm{g}}}\\) where \u03c1 is the density of cork. (Ignore damping due to viscosity of the liquid).
                                            \nAnswer:
                                            \nLet x be the depression created.
                                            \nExcess upthrust caused is, Ue<\/sub>= g\\(\\rho_{\\mathrm{e}}\\) Ax
                                            \nRestoring force = U = gA\\(\\rho_{\\mathrm{e}}\\) x
                                            \nmass = m = Ah\u03c1
                                            \nNow F = m\u03b1;\u00a0 \u03b1 = acceleration
                                            \nAh\u03c1\u03b1 = -gA\\(\\rho_{\\mathrm{e}}\\) x<\/p>\n

                                            \"1st<\/p>\n

                                            Question 4.
                                            \nIf this earth were a homogenous sphere and a straight hole bored in it through its centre. Show that if a body were dropped into the hole it would execute a SHM. Also, find its time period.
                                            \nAnswer:
                                            \nLet mass of body dropped = m
                                            \nmass of earth = M
                                            \nradius of earth = R
                                            \nmg = \\(=\\frac{\\mathrm{GMm}}{\\mathrm{R}^{2}}\\) \u21d2\u00a0 g = \\(\\frac{\\mathrm{GM}}{\\mathrm{R}^{2}}\\)
                                            \n\u21d2\u00a0 \\(\\mathrm{g}=\\frac{\\mathrm{G}}{\\mathrm{R}^{2}} \\frac{4}{3} \\pi \\mathrm{R}^{2} \\rho=\\frac{4 \\pi \\mathrm{GR} \\rho}{3}\\)
                                            \nwhere \u03c1 is mean density of earth
                                            \n\"1st
                                            \nLet R be the depth of the body fallen from centre then g at the point is
                                            \n\\(\\mathrm{g}^{\\prime}=\\frac{\\mathrm{Gm}^{\\prime}}{(\\mathrm{R}-\\mathrm{d})^{2}}\\)
                                            \nwhere m’ is mass of sphere of radius (R – d)
                                            \n\"1st
                                            \nHence acceleration is proportional to displacement and in opposite direction go on SHM.
                                            \n\"1st<\/p>\n","protected":false},"excerpt":{"rendered":"

                                            You can Download Chapter 14 Oscillations Questions and Answers, Notes, 1st PUC Physics Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka 1st PUC Physics Question Bank Chapter 14 Oscillations 1st PUC Physics Oscillations Textbook Questions and Answers Question 1. Which of …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[81],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/9204"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=9204"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/9204\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=9204"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=9204"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=9204"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}