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Simple Harmonic Motion Test - 5

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Simple Harmonic Motion Test - 5
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  • Question 1
    1 / -0

    A small sphere carrying a charge ‘q’ is hanging in between two parallel plates by a string of length L. Time period of pendulum is T0. When parallel plates are charged, the electric field between the plates is E and time period changes to T. The ratio T/T0 is equal to 

    Solution

    Compare this system with a simple pendulum oscillating only under gravity. Calculate effective acceleration in a vertically downward direction.

    Step 1: Draw a diagram as follows.

    Step2: Calculate effective acceleration g'.

    Net downward force mg' = mg+QE

    Step3 : Calculate timeperiod by using the effective acceleration.

     

  • Question 2
    1 / -0

    A simple pendulum has a time period T1 when on the earth’s surface and T2 when taken to a height R above the earth’s surface, where R is the radius of the earth. The value of T2/T1 is:

    Solution

    If acceleration due to gravity is at the surface of Earth, then at height R its value becomes

     

  • Question 3
    1 / -0

    A particle executes linear simple harmonic motion with an amplitude of of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity  is equal to that of its acceleration. Then, its time period in seconds is

    Solution

    magnitude of velocity of particle when it is at displacement x from mean position

    Also, magnitude of acceleration of particle in SHM

    = ω2x

    Given, when x=2cm

    |v|=|a|

    ⇒Angular velocity ω = √5/2

    So, Time period of motion 

     

  • Question 4
    1 / -0

    When two displacements represented by y1=asin(ωt) and y2=bcos(ωt) are superimposed,the motion is -

    Solution

    Given,

    y= a sin ωt

    y= b cos ωt = b sin (ωt + π/2)

    The resultant displacement is given by;

    Hence, the motion of superimposed wave is simple harmonic with amplitude √a2+b2 .

     

  • Question 5
    1 / -0

    The displacement of a particle along the r-axis is given by x = asin2ωt,t. The motion of the particle corresponds to 

    Solution

    For the given displacement x a sin2ω/t,

    a ∝-x is not satisfied

    Hence, the motion of the particle is non simple harmonic motion.

    Note : The given motion is a periodic motion with a time period

     

  • Question 6
    1 / -0

    The period of oscillation of a mass M suspended from a spring of negligible mass is T. If along with it another mass M is also suspended, the period of oscillation will now be

    Solution

     

  • Question 7
    1 / -0

    A body performs simple harmonic motion about x=0 with an amplitude a and a time period T. The speed of the body at x = a/2 will be:

    Solution

     

  • Question 8
    1 / -0

    Which one of the following equations of motion represents simple harmonic motion?

    where k, k0, k1 and αα are all positive.

    Solution

    Acceleration ∝−(displacement).

    Here, y=x+a

    ∴ acceleration =-k(x+a)

     

  • Question 9
    1 / -0

    Two simple harmonic motions of angular frequency 100 rad s -1 and 1000 rad s−1 have the same displacement amplitude. The ratio of their maximum acceleration will be

    Solution

    Acceleration of simple harmonic motion is

     

  • Question 10
    1 / -0

    A point performs simple harmonic oscillation of period T and the equation of motion is given by x= a sin (ωt +π/6). After the elapse of what fraction of the time period the velocity of the point will be equal to half to its maximum velocity?

    Solution

    Velocity is the time derivation of displacement. Writing the given equation of a point performing SHM

    Differentiating Eq. (i),w.r.t. time, we obtain

    It is given that v = aω/2, so that 

    Thus, at T/12 velocity of the point will be equal to half of its  maximum velocity.

     

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