Physics Part Ii Chapter 14 Oscillations
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    NCERT Solution For Class 11 Physics Physics Part Ii

    Oscillations Here is the CBSE Physics Chapter 14 for Class 11 students. Summary and detailed explanation of the lesson, including the definitions of difficult words. All of the exercises and questions and answers from the lesson's back end have been completed. NCERT Solutions for Class 11 Physics Oscillations Chapter 14 NCERT Solutions for Class 11 Physics Oscillations Chapter 14 The following is a summary in Hindi and English for the academic year 2021-2022. You can save these solutions to your computer or use the Class 11 Physics.

    Question 14
    CBSEENPH11020484

    The bob of a simple pendulum executes simple harmonic motion in water with a period t, while the period of oscillation of the bob is t0 in the air. Neglecting frictional force of water and given that the density of the bob is (4/3) x 1000 ms-1 . What relationship between t and t0 is true?

    • t = t0

    • t = t0/2

    • t = 2t0

    • t = 4t0

    Solution

    C.

    t = 2t0

    The time period of simple pendulum in air
    straight T space equals space straight t subscript 0 space equals space 2 straight pi space square root of 1 over straight g end root space...... space left parenthesis straight i right parenthesis
    being the length of simple pendulum, In water, effective weight of bob
    w' = weight of bob in air - upthrust
    ⇒ ρVgeff = mg-m'g = ρVg-ρ'Vg = (ρ-ρ') where ρ = density of bob, ρ′ = density of water
    straight g subscript eff space equals space open parentheses fraction numerator straight rho minus straight rho apostrophe over denominator straight rho end fraction close parentheses space straight g
space equals space open parentheses 1 minus fraction numerator straight rho apostrophe over denominator straight rho end fraction close parentheses straight g
therefore space
space straight t equals space 2 straight pi space square root of open square brackets fraction numerator 1 over denominator open parentheses 1 minus begin display style fraction numerator straight rho apostrophe over denominator straight p end fraction end style close parentheses end fraction close square brackets end root space........ space left parenthesis ii right parenthesis
Thus comma space straight t over straight t subscript straight o space equals space square root of open square brackets fraction numerator 1 over denominator 1 minus begin display style fraction numerator straight rho apostrophe over denominator straight rho end fraction end style end fraction close square brackets end root
space equals space square root of open parentheses fraction numerator 1 over denominator 1 minus begin display style fraction numerator 1000 over denominator left parenthesis 4 divided by 3 right parenthesis space straight x space 1000 end fraction end style end fraction close parentheses end root space equals space square root of open parentheses fraction numerator 4 over denominator 4 minus 3 end fraction close parentheses end root

space equals space 2 rightwards double arrow space straight t space space equals 2 straight t subscript 0

    Question 15
    CBSEENPH11020485

    A particle at the end of a spring executes simple harmonic motion with a period t1, while the corresponding period for another spring is t2. If the period of oscillation with the two springs in series is t, then

    • T = t1 + t2

    • straight T squared space equals space straight t subscript 1 superscript 2 space plus space straight t subscript 2 superscript 2
    • space straight T to the power of negative 1 end exponent space equals straight t subscript 1 superscript negative 1 end superscript space plus straight t subscript 2 superscript negative 1 end superscript
    • space straight T to the power of negative 1 end exponent space equals straight t subscript 1 superscript negative 2 end superscript space plus straight t subscript 2 superscript negative 2 end superscript

    Solution

    B.

    straight T squared space equals space straight t subscript 1 superscript 2 space plus space straight t subscript 2 superscript 2

    Time period of spring
    straight T space equals space 2 straight pi square root of straight M over straight k end root
    k, being the force constant of spring. For first spring and for second spring
    we have,
    straight T space equals space 2 straight pi square root of straight m over straight k subscript 1 end root space space space....... space left parenthesis straight i right parenthesis

straight T space equals space 2 straight pi square root of straight m over straight k subscript 2 end root space......... space left parenthesis ii right parenthesis
    The effective force constant in their series combination is
    straight k space equals fraction numerator straight k subscript 1 straight k subscript 2 over denominator straight k subscript 1 plus straight k subscript 2 end fraction
therefore space Time space period space of space combination

straight T space equals space 2 straight pi space square root of open square brackets fraction numerator straight m left parenthesis straight k subscript 1 plus straight k subscript 2 right parenthesis over denominator straight k subscript 1 straight k subscript 2 end fraction close square brackets end root
straight T squared space equals space fraction numerator 4 straight pi squared straight m space left parenthesis straight k subscript 1 plus straight k subscript 2 right parenthesis over denominator straight k subscript 1 straight k subscript 2 end fraction space...... space left parenthesis iii right parenthesis
From space equations space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis comma space we space obtain
straight t subscript 1 superscript 2 space plus space straight t subscript 2 superscript 2 space equals space 4 straight pi squared space open parentheses straight m over straight k subscript 1 space plus straight m over straight k subscript 2 close parentheses
rightwards double arrow straight t subscript 1 superscript 2 space plus space straight t subscript 2 superscript 2 space equals space 4 straight pi squared straight m space open parentheses 1 over straight k subscript 1 plus 1 over straight k subscript 2 close parentheses
straight t subscript 1 superscript 2 space plus straight t subscript 2 superscript 2 space equals space fraction numerator 4 straight pi squared straight m space left parenthesis straight k subscript 1 plus straight k subscript 2 right parenthesis over denominator straight k subscript 1 straight k subscript 2 end fraction
straight t subscript 1 superscript 2 space plus straight t subscript 2 superscript 2 space equals space straight T squared

    Question 16
    CBSEENPH11020486

    The total energy of particle, executing simple harmonic motion is

    • ∝ x

    • ∝ x2

    • independent of x

    • ∝ x1/2

    Solution

    C.

    independent of x

    In simple harmonic motion when a particle is displaced to a position from its mean position, then its kinetic energy gets converted into potential energy and vice-versa. Hence, total energy of a particle remains constant or the total energy in simple harmonic motion does not depend on displacement x.

    Question 17
    CBSEENPH11020488

    A particle of mass m is attached to a spring (of spring constant k) and has a natural angular frequency ω0. An external force F(t) proportional to cosωt (ω≠ω0) is applied to the oscillator. The time displacement of the oscillator will be proportional to

    • fraction numerator straight m over denominator straight omega subscript 0 superscript 2 minus straight omega squared end fraction
    • fraction numerator 1 over denominator straight m left parenthesis straight omega subscript 0 superscript 2 minus straight omega squared right parenthesis end fraction
    • fraction numerator 1 over denominator straight m left parenthesis straight omega subscript 0 superscript 2 plus straight omega squared right parenthesis end fraction
    • fraction numerator straight m over denominator straight omega subscript 0 superscript 2 plus straight omega squared end fraction

    Solution

    B.

    fraction numerator 1 over denominator straight m left parenthesis straight omega subscript 0 superscript 2 minus straight omega squared right parenthesis end fraction

    Initial angular velocity of particle = ω0 and at any instant t, angular velocity = ω Therefore, for a displacement x, the resultant acceleration
    straight f equals space left parenthesis straight omega subscript 0 superscript 2 space minus straight omega squared right parenthesis straight x space space space..... left parenthesis straight i right parenthesis
External space force
straight F space equals space straight m left parenthesis straight f equals space left parenthesis straight omega subscript 0 superscript 2 space minus straight omega squared right parenthesis space straight x....... left parenthesis ii right parenthesis
Since comma space straight F proportional to space cosωt space left parenthesis given right parenthesis
therefore space
From space eq. space left parenthesis ii right parenthesis space straight m space left parenthesis straight omega subscript 0 superscript 2 space minus straight omega squared right parenthesis space straight x space proportional to space cosωt space.... space left parenthesis iii right parenthesis
    Now, equation of simple harmonic motion
    x = A sin (ωt + φ) .......... (iv)
    at t = 0 ; x = A
    ∴ A = A sin( 0 + φ )
    ⇒  φ =π/2
    therefore space straight A space sin space open parentheses ωt space plus straight pi over 2 close parentheses space equals space straight A space cot space ωt ..........(v)
    Hence, from equations (iii) and (v), we finally get
    straight m left parenthesis straight omega subscript 0 superscript 2 minus straight omega squared right parenthesis space straight A space cos space ωt space proportional to space cos space ωt
straight A proportional to space fraction numerator 1 over denominator straight m left parenthesis straight omega subscript 0 superscript 2 minus straight omega squared right parenthesis end fraction

    Sponsor Area

    Question 18
    CBSEENPH11020489

    In forced oscillation of a particle the amplitude is maximum for a frequency ω1 of the force, while the energy is maximum for a frequency ω2 of the force, then

    • ω1 = ω2

    • ω1 > ω2

    • ω1 < ω2 when damping is small and ω1 > ω2 when damping is large

    • ω1 < ω2

    Solution

    A.

    ω1 = ω2

    For amplitude of oscillation and energy to be maximum, frequency of force must be equal to the initial frequency and this is only possible in case of resonance. In resonance state ω1 = ω2

    Question 22
    CBSEENPH11020835

    A tuning fork is used to produce resonance in a glass tube. The length of the air column in this tube can be adjusted by a variable piston. At room temperature of 27ºC two successive resonances are produced at 20 cm and 73 cm of column length. If the frequency of the tuning fork is 320 Hz, the velocity of sound in air at 27ºC is

    • 330 m/s

    • 339 m/s

    • 300 m/s

    • 350 m/s

    Solution

    B.

    339 m/s

    Two successive resonance is produced at 20 cm and 73 cm of column length

     λ2 = (73-20) x 10-2 m λ = 2 x (73-20) x 10-2Velocity of sound, v = nλ= 2 x 320 [ 73 - 20] x 20-2= 339.2 ms-1

    Question 23
    CBSEENPH11020836

    A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 m/s2 at a distance of 5 m from the mean position. The time period of oscillation is

    • 2πs

    • π s

    • 1s

    • 2s

    Solution

    B.

    π s

    Acceleration, a = 20 m/s2 , and displacement, y = 5
    |a| =ω2y 20 = ω2(5) ω = 2 rad/sTime period of pendulum,T = 2πω = 2π2 = πs

    Question 24
    CBSEENPH11020860

    A particular of mass m is executing oscillation about the origin on X- axis. Its potential energy is V(x) = K |x|3. Where K is a positive constant. If the amplitude of oscillation is a, then its time period T is proportional to

    • 1a

    • a

    • a

    • a3/2

    Solution

    A.

    1a

    Given, V = kx3

    Total energy (E) = Maximum potential energy

    = ka3

    From conservation of energy

    12 m dxdt2 + kx2 = ka3 12m dxdt2 = k(a3 -x3) dxdt = 2km(a3-x3)dxa3-x3 = 2kmdtOr integrating0a dxa3 - x3 = 2km0T/4dtor 0a dxa3-x3 = 2kmT4Let x = a sin θ  dx = a cos θ  0π/2 a cos θ a3 -(a sin θ)3 = 2Km T4

    Question 25
    CBSEENPH11020874

    The displacement of a particle along the x-axis is given by x = a sin2 ωt. The motion of the particle corresponds to 

    • Simple harmonic motion of frequency ω/π

    • Simple harmonic motion of frequency  3ω/2π

    • non-simple harmonic motion

    • Simple harmonic motion of frequency ω/2π

    Solution

    C.

    non-simple harmonic motion

    Given X = a sin2 ωt

    X = a 1- cos 2 ωt2 cos 2θ = 1 -2 sin2 θor X = a2 - a cos 2ωt2Now, V = dxdt =  sin 2ωtand a = dvdt = 2 2 cos 2ωt

    Here, a is not directly proportional  to (-x) which is a condition for SHM.

    Question 26
    CBSEENPH11026108

    A simple pendulum of length l has a maximum angular displacement θ. The maximum kinetic energy of the bob is

    • mgl (1-cosθ)

    • 0.5 mgl

    • mgl

    • 0.5 mgl

    Solution

    A.

    mgl (1-cosθ)

    Here length of the pendulum = l

    Maximum angular displacement =θ

    Mass of the bob = m

    Now height of the bob at maximum angular displacement is

    h = l - l (1-cosθ)

    Also at the end of displacement kinetic energy of the bob = potential energy of the bob

    mgh = mgl (1-cosθ)

    hence kinetic energy of the bob = mgl (1-cosθ)

    Question 27
    CBSEENPH11026120

    Equations of motion in the same direction are given by

    y1 =2α sinωt -kx   y2 =2α sinωt -kx-θ 

    The amplitude of the medium particle will be

    • 2α cosθ

    • 2α cosθ

    • 4α cosθ2

    • 2α cosθ2

    Solution

    C.

    4α cosθ2

    The given equations of motion are

    y1=2 α  sin ωt - kxy2 =2 α  sin ωt  -kx -θNow the resultant eqation of wave is given byy= y1 + y2y =2α sin ωt - kx  + 2α sin ωt -kx-θy =2α 2 sin ωt - kx +ωt -kx -θ2                                  ×2cos ωt - kx -wt -kx -θ2 y=4α cosθ2 sin ωt -kx -θ2             .....(1)Now compairing equation (1) withy = A sin ωt - kxResultant amplitude isA = 4 α cosθ2 

    Question 28
    CBSEENPH11026134

    If the length of a pendulum is made 9 times and mass of the bob is made 4 times, then the value of time period becomes :

    • 3 T

    • 3/2T

    • 4T

    • 2T

    Solution

    A.

    3 T

    The time for one complete cycle, a left swing, is called the period. The period depends on length of the period. The period depends on the length of the pendulum and also to a slight degree on the amplitude, the width of the pendulum's swing.

    Time period  = 2πlgi.e time period does not depend on the mass of the bob.T  lT1T2 =l1g1TT2= l9lT2 = 3T

    Question 29
    CBSEENPH11026137

    A pendulum is undergoing SHM with frequency f What is the frequency of its kinetic energy ?

    • f2

    • 2 f

    • 3 f

    • 4 f

    Solution

    B.

    2 f

    Suppose  A particle in s.H.M follows the equation

    X = A sin ωtdXdt = V =A ω cosωtSo, kinetic energy = 12  mv2                                =12m A2 ω2  cos2ωtNow if a particlewould have been moving following such equation i.eY =12m A2 ω2 cos2ωtY = 14 m A2 ω2 2 cos2ωtY= 14 m A2 ω2 sin2ωt +1Y -  14m A2 ω2   =  14m A2 ω2 sin 2ωty= 14m A2 ω2 sin 2ωt                                             since y=Y -14m A2ω2= constant14m A2ω2 - Amplitudesin 2ω t - has been doubled so frequency is doubled as well 

    Question 30
    CBSEENPH11026151

    Two equal -ve charges -q are fixed at the point (O, a) and (O, a) on the y-axis. A positive charge Q is released from rest at the point (2a, 0) on the x-axis. The charge will :

    • execute SHM about the origin

    • move to the origin and remain at rest

    • move to infinity

    • execute oscillatory but not SHM

    Solution

    D.

    execute oscillatory but not SHM

    Component of force on charge +Q at P, along the x-axis,

    Fx =2Qq4πεo a2 + x2 cosθ     =2Qq4πεo a2 + x2 × xa2 + x2 Fx  = 2Qqx4πεoa2 + x232

    Which is directly proportional to x. So, motion is an oscillatory but not SHM.

    Question 31
    CBSEENPH11026157

    The circular motion of a particle with constant speed is

    • simple harmonic but not periodic

    • periodic and simple harmonic

    • neither periodic nor simple harmonic

    • periodic but not simple harmonic

    Solution

    D.

    periodic but not simple harmonic

    In a circular motion particle repeats after equal intervals of time. So, particle motion on a circular path is periodic but not simple harmonic as it does not execute to and fro motion about a fixed point.

    Question 32
    CBSEENPH11026158

    A particle executing simple harmonic motion of amplitude 5 cm has maximum speed of 31.4 cm/s. The frequency of its oscillation is

    • 3 Hz

    • 2 Hz

    • 4 Hz

    • 1 Hz

    Solution

    D.

    1 Hz

    The maximum speed of a particle executing SHM is

    umax =  = a (2πn)n = umax2πaHere, umax = 31.4 cm/s, a = 5 cm

    substituting the given values, we have

    n = 3.142 × 3.14 × 5 n = 1 Hz

    Question 33
    CBSEENPH11026175

    A particle executes simple harmonic oscillation with an amplitude α . The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is

    • T4

    • T8

    • T12

    • T2

    Solution

    C.

    T12

    Let displacement equation of particle executing SHM is

    y = α sinωt

    As the particle travels half of the amplitude from the equilibrium position, so

    y = α2 α2 = α sinωtsin ωt = 12 = sin π6     ωt  = 12 = sin π6     ωt = π6      t = π6ω

     t = π6 2πTt = T12

    Hence, the particle travels half of the amplitude from the equilibrium in T12s.

    Question 34
    CBSEENPH11026207

    Given that y = A sin  2πλ ct - x  , where y and x are measured in metre. Which of the following statements is true?

    • The unit of λ is same as that of x and A

    • The unit of λ is same as that of x but not of A

    • The unit of c is same as that of 2πλ

    • The unit of ( ct - x ) is same as that of 2πλ

    Solution

    A.

    The unit of λ is same as that of x and A

    Here 2πλ ct - x is dimensionless

    Hence, ctλ is also dimensionless and unit of ct is the same as that of x. Therefore unit of λ is same as that of x.

    Also unit of y is same as that of A, which is also the unit of x.

    Question 35
    CBSEENPH11026293

    Two simple pendulums of length 0.5 m and 20 m respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has
    completed x oscillations, where k is

    • 1

    • 3

    • 2

    • 5

    Solution

    C.

    2

    Time period of simple pendulum

    For 1st pendulum T12π5g

                                   = 2π209.8

                                    = 20 s

    For IInd pendulum T22π 20g

                                     = 2π 209.8

                                     = 2 20s

    Suppose the time will be in same phase is t then

                  1t = 1T1 - 1T2

                       = 120 - 1220

                       = 12 20

                  t = 220

    Hence the number of oscillation of simple pendulum of shorter wavelength

            (x) = 22020

             (x)  =  2             

    Question 36
    CBSEENPH11026308

    A mass of 1 kg is suspended from a spring of force constant 400 N, executing SHM total energy of the body is 2J, then maximum acceleration of the spring will be

    • 4 m/s2

    • 40 m/s2

    • 200 m/s2

    • 400 m/s2

    Solution

    B.

    40 m/s2

    Oscillations due to spring

               T = 2π mR

    Where T is the time period

              k is spring constant

              m is mass attached to spring

             T = 2πmR

                  = 2π 1400

                  = 2π20

        Frequency n = 202 π

         ω = 2π n

             = 2π × 202π

           ω = 20

      Total energy

               E = 122 r2

                2 = 12m ω2r2

                r = 110 m

       Acceleration = ω2 r

                          = 400 × 110

                           = 40 m/s2

    Question 37
    CBSEENPH11026309

    Vibrations of rope tied by two rigid ends shown by equation y = cos 2πt sin2πx, then minimum length of the rope will be

    • 1 m

    • 12m

    • 5 m

    • 2πm

    Solution

    B.

    12m

    Equation of vibration

            y = 1( cos2πt sin2πx )

               = 1 cos2π sinπ x12

    Comparing this equation with

           y = a cos 2π sin π xλ

     Hence fro above equation

                a = 1   

    Minimum length of string

              =  12 a 

              = 12 × 1

    Mininmum length of string

               = 12 m

    Question 38
    CBSEENPH11026313

    The angular amplitude of a simple pendulum is θ0. The maximum tension in its string will be

    • mg (1 - θ0)

    • mg (1 + θ0 )

    • mg 1- θ02

    • mg 1 + θ02

    Solution

    D.

    mg 1 + θ02

    The simple pendulum at angular amplitude θo is shown in the figure

             Tmaxmg + mv2l            ....(i)

    When bob of pendulum comes from A to B, it covers a vertical distance h.

       

    ∴       cosθ0 = l - hl

    ⇒        h = l ( l - cos θ )                 .....(ii) 

    Also during A to B, potential energy of bob converts into kinetic energy

    ∴        mgh = 12mv2

    ∴            v = 2 g h                   ....... (iii)

    Thus, using Eqs. (i), (ii) and (iii) we obtain

        Tmax = mg + 2mgl l  1 - cosθo 

                = mg + 2mg 1 - 1 + θ022

                              (By using binomial theorem)

                = mg 1 + θ02

    Question 39
    CBSEENPH11026332

    A particle is subjected to two simple harmonic motions along X-axis while other is along a line making angle 45° with the X-axis. The two motions are given by   x = x0 sin ωt  and  s =s0 sin ωt. The amplitude of resultant motion is 

    • x0 + s0 + 2x0s0

    • x02 + s02

    • x02 - s02 + 2x0 s0

    • x02 + s02 + 2 x0 s012

    Solution

    D.

    x02 + s02 + 2 x0 s012

    Two SHMs are separated at angle 45° So, resultant amplitude is

             ARs0 + x02 cos2 45o 2 + x0 sin 45o2

                  =  s02 + x02 cos2 45o+ 2x0 s0 cos45o + x02 sin2 45o 

     Since sin 45o12⇒ sin2 45o12

               cos 45o12⇒ cos2 45o12 

                ARs02 + x02 12 + 2x0 s0 12 + x02 12    

     ⇒        ARs02 +x02 + 2 x0 s0 

    Sponsor Area

    Question 40
    CBSEENPH11026358

    A pan with a set of weights is attached to a light spring. The period of vertical oscillation is 0.5s. When some additional weights are put in pan, then the period of oscillations increases by 0.1 s. The extension caused by the additional weight is

    • 5.5 cm

    • 3.8 cm

    • 2.7 cm

    • 1.3 cm

    Solution

    C.

    2.7 cm

    Given:- T = 0.5 s

    We know that,

    The time period for oscillation is

                T = 2π mk

    k = It is spring constant

    m = mass 

    ⇒          T = 0.5

    ⇒          0.5 = 2π mk

    ⇒           14π = mk

    On squaring on both sides, we get

    ⇒            mk = 116 π2                      ....(i)

    ⇒         T + 0.1 = 2π m + m'k

    ⇒     0.5 + 0.1 = 2π m + m'k

    ⇒       0.62π2 = m + m'k 

    ⇒                  = mk+m'k

    ⇒         0.624π2 = 116 π2 + m'k                   using eq.(i)

    ⇒           0.6 24π2 - 116 π2 = m'k 

    ⇒            m'k = 1.44 - 116 π2

    ⇒              m'k = 0.4416 π2                          ....(ii)

                   m' g = kx

                   x = m'kg

    ⇒                 = 0.4416 π2 × 10

    ⇒            x = 0.0279

    ⇒            x = 2.79 × 10-2 m

    ×            x = 2.7 cm

    Question 41
    CBSEENPH11026391

    A particle is describing simple harmonic motion. If its velocities are v1 and v2 when the displacements from the mean position are y1 and y2 respectively, then its time period is

    • 2π y12 + y22v12 + v22

    • 2π v22 - v12y12 - y22

    • 2π v12 + v22y12 + y22

    • 2π y12  - y22v22 - v12

    Solution

    D.

    2π y12  - y22v22 - v12

    In simple harmonic motion,

       velocity 

             v = ω A2 - y2

    ∴       v1ω A2 - y12

    ⇒       v12 = ω2A2 - ω2 y12          .....(i)

    and     v2 = ω A2 - y22

    ⇒        v22 = ω2A2 - ω2y22          ....(ii)

    Solving equations (i) and (ii), we get

             v22 - v12 = ω2 y12 - y22

              ω = v12 - y22 v22 - v12

    Time period is given by,

    ⇒         T = 2πω

    ⇒        T = 2π y12 - y22v12 - v12

    Question 43
    CBSEENPH11026414

    A body of mass 60 kg is suspended by means of three strings P Q and R as shown in the figure is in equilibrium. The tension in the string P is

      

    • 130.9 g N

    • 60 g N

    • 50 g N

    • 103.9 g N

    Solution

    D.

    103.9 g N

           

           R cos 60o = Q = 60 g

    ⇒           R = 60 gcos 60o                    cos60o = 12

                   P = R sin60o

                       = 120 g × 32

     ⇒           P = 103.9 g N

    Question 44
    CBSEENPH11026415

    The angular amplitude of simple pendulum is θo. The maximum tension in its string will be

    • mg ( 1 - θ0 )

    • mg ( 1 + cosθ)

    • mg ( 1 - cosθ02 )

    • mg ( 1 + cosθ02 )

    Solution

    D.

    mg ( 1 + cosθ02 )

    Maximum tension in the string is

                   Tmax =  mg + m v2l 

                           =  mg + 2 mg ll ( 1 - cosθ0 )

                            = mg + 2mg 1 - 1 + θ022                        ( since θ0 is small )

                   Tmax = mg ( 1 + θ)     

    Question 45
    CBSEENPH11026425

    The displacement of a particle executing SHM is given by y = 0.25 sin200t cm. The maximum speed of the particle is

    • 200 cm s-1

    • 100 cm s-1

    • 50 cm s-1

    • 5.25 cm s-1

    Solution

    C.

    50 cm s-1

    Given:- displacement of particle executing SHM  

           y = 0.25 sin 200t

    speed = dydt

              = 0.25 × 200 cos 200t

    i.e cos 200t = 1

    Max speed= 0.25 x 200

                   = 50 cm s-1

    Question 46
    CBSEENPH11026427

    A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes simple harmonic oscillations with a time period T. If the mass is increased by m, then the time period becomes 54 T. The ratio of mM is

    • 9/16

    • 5/4

    • 25/16

    • 4/5

    Solution

    A.

    9/16

    The time period of oscillation is 

              T = 2π MK

    M = mass attached to spring

    and       T1 = 2π M + mK

    ⇒          T1T = M + mM

    ⇒           54 = 1 + mM12                    T1 = 54T

    ⇒             mM = 2516 - 1

    ⇒               mM = 916

    Question 47
    CBSEENPH11026428

    A wave is represented by the equation y= 0.5 sin(10t - x) metre It is a travelling wave propagating along +x direction with velocity

    • 10 m s-1

    • 20 m s-1

    • 5 m s-1

    • None of these

    Solution

    A.

    10 m s-1

    Compare the given eqution y = 0.5 sin (10t - x) with the standard form

              y = r sin 2π tT + 2π xλ

              2πT = 10

               2πλ = 1

     ⇒        v = λt

                   = 101

     ⇒        v = 10 m/s

    Question 48
    CBSEENPH11026430

    The frequency of oscillations of a mass m connected horizontally by a spring of spring constant k is 4 Hz. When the spring is replaced by two identical spring as shown in figure. Then the effective frequency is

        

    • 2

    • 1.5

    • 1.31

    • 22

    Solution

    D.

    22

    If one spring is connected then frequency = 4Hz

              4 = 12πkm                       ....(i)

    If two springs are connected in series, then frequency

              v = 12πk'm                      ....(ii)

          k' = k2

    From equation (i) and (ii)

          4v = 12πkm12πk2m

          4v = 2

         v = 2 2 Hz

    Question 49
    CBSEENPH11026436

    Assertion:  The graph of potential energy and kinetic energy of a particle in SHM with respect to position is a parabola. 

    Reason:  Potential energy and kinetic energy do not vary linearly with position.

    • If both assertion and reason are true and reason is the correct explanation of assertion.

    • If both assertion and reason are true but reason is not the correct explanation of assertion.

    • If assertion is true but reason is false.

    • If both assertion and reason are false.

    Solution

    B.

    If both assertion and reason are true but reason is not the correct explanation of assertion.

    Both assertion and reason are true but reason is not the correct explanation of assertion.

                  P. E = 12 m ω2 y2

    and         K.E = 12 m ω2 ( α2 - y2 )

    The variation of both potential energy and kinetic energy is proportional to y2

    Therefore displacement versus P.E. or K.E. graph is a parabola ( because parabolic equation is  z = 4au2 )

    Question 50
    CBSEENPH11026448

    Motion of an oscillating liquid in a U tube is

    • periodic but not simple harmonic.

    • non-periodic.

    • simple harmonic and time period is independent of the density of the liquid.

    • simple harmonic and time period is directly proportional to the density of the liquid.

    Solution

    C.

    simple harmonic and time period is independent of the density of the liquid.

    Motion of an oscillating liquid column in a U tube is SHM with period

                    T = 2π hg

    where h is the height of liquid column in one arm of U tube in equilibrium position of
    liquid. Therefore, T is independent of density of liquid.

    Question 51
    CBSEENPH11026464

    The displacement x of a particle varies with time t as, x = ae-αt +b e βt where a, b α and β are positive constants. The velocity of the particle will

    • go on decreasing with time

    • be independent of α and β

    • drop to zero when α = β

    • go on increasing with time

    Solution

    D.

    go on increasing with time

    x = a e αt +b e βtSo velocity  v = dXdt   =-a αe-αt + b βe βeν = A + BWhere A = -a αe-αt , B =b βe βr

    The value of term A = a e-αt decreases and of term B = b βeβe increases with increase in time. As a result, velocity goes on increasing with time.

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