# 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)   $±\frac{A}{\sqrt{2}}$

Subtopic:  Energy of SHM |
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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) $U=KX$ 


Subtopic:  Energy of SHM |
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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)   $\left({a}^{2}-{x}^{2}\right)/{X}^{2}$

Subtopic:  Energy of SHM |
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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

Subtopic:  Energy of SHM |
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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

Subtopic:  Energy of SHM |
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The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (where E is the total energy)

(1)    $\frac{1}{8}E$

(2)        $\frac{1}{4}E$

(3)    $\frac{1}{2}E$

(4)        $\frac{2}{3}E$

Subtopic:  Energy of SHM |
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A body executes simple harmonic motion. The potential energy (P.E.), the kinetic energy (K.E.) and total energy (T.E.) are measured as a function of displacement x. Which of the following statements is true ?

(1)  P.E. is maximum when x = 0

(2)  K.E. is maximum when x = 0

(3)  T.E. is zero when x = 0

(4)  K.E. is maximum when x is maximum

Subtopic:  Energy of SHM |
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­­A man measures the period of a simple pendulum inside a stationary lift and finds it to be T sec. If the lift accelerates upwards with an acceleration $\mathrm{g}}{4}$ , then the period of the pendulum will be

(1) T

(2) $\frac{\mathrm{T}}{4}$

(3) $\frac{2\mathrm{T}}{\sqrt{5}}$

(4) $2\mathrm{T}\sqrt{5}$

Subtopic:  Angular SHM |
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The total energy of a particle, executing simple harmonic motion is

(1)

(2)

(3)   Independent of x

(4)

Subtopic:  Energy of SHM |
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The bob of a pendulum of length l is pulled aside from its equilibrium position through an angle $\theta$ and then released. The bob will then pass through its equilibrium position with a speed v, where v equals

(1) $\sqrt{2\mathrm{gl}\left(1-\mathrm{sin\theta }\right)}$

(2) $\sqrt{2\mathrm{gl}\left(1+\mathrm{cos\theta }\right)}$

(3) $\sqrt{2\mathrm{gl}\left(1-\mathrm{cos\theta }\right)}$

(4) $\sqrt{2\mathrm{gl}\left(1+\mathrm{sin\theta }\right)}$

Subtopic:  Angular SHM |
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