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#10-Solved Examples 3
(Physics) > Oscillations

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Related Practice Questions :

A particle executes simple harmonic motion along a straight line with an amplitude A. The potential energy is maximum when the displacement is

(1)     $±A$

(2)    Zero

(3)     $±\frac{A}{2}$

(4)   $±\frac{A}{\sqrt{2}}$

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The potential energy of a particle with displacement X depends as U(X). The motion is simple harmonic, when (K is a positive constant)

(1) $U=\frac{K{X}^{2}}{2}$

(2) $U=K{X}^{2}$

(3)  $U=K$

(4) $U=KX$ 


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The angular velocity and the amplitude of a simple pendulum is $\omega$ and a respectively. At a displacement X from the mean position if its kinetic energy is T and potential energy is V, then the ratio of T to V is

(1)     ${X}^{2}{\omega }^{2}\left({a}^{2}-{X}^{2}{\omega }^{2}\right)$

(2)    ${X}^{2}/\left({a}^{2}-{x}^{2}\right)$

(3)   $\left({a}^{2}-{X}^{2}{\omega }^{2}\right)/{X}^{2}{\omega }^{2}$

(4)   $\left({a}^{2}-{x}^{2}\right)/{X}^{2}$

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A particle is executing simple harmonic motion with frequency f. The frequency at which its kinetic energy changes into potential energy, will be:

1.   f/2

2.  f

3.   2 f

4.  4 f

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There is a body having mass m and performing S.H.M. with amplitude a. There is a restoring force ,$F=-Kx$ where x is the displacement. The total energy of body depends upon -

(1)   K, x

(2)  K, a

(3)   K, a, x

(4)  K, a, v

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