(C) The potential energy $U$ of a particle executing simple harmonic motion is given by $U = \frac{1}{2} k x^2$,where $k$ is the force constant and $x

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(C) The potential energy $U$ of a particle executing simple harmonic motion is given by $U = \frac{1}{2} k x^2$,where $k$ is the force constant and $x$ is the displacement from the mean position.For the potential energy to be maximum,the displacement $x$ must be at its maximum value.In simple harmonic motion,the maximum displacement from the mean position is equal to the amplitude $A$.Therefore,the potential energy is maximum at the extreme positions,which are $x = \pm A$.

Class 11 Physics Oscillations Energy of Simple Harmonic Motion

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$A$ particle executing simple harmonic motion along a straight line with an amplitude $A$,attains maximum potential energy when its displacement from the mean position equals

AP EAMCET 2021 All AP EAMCET

A
$0$
B
$\pm \frac{A}{\sqrt{2}}$
C
$\pm A$
D
$\pm \frac{A}{2}$

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$A$ particle executing simple harmonic motion along a straight line with an amplitude $A$,attains maximum potential energy when its displacement from the mean position equals

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$A$ particle of mass $10 \ g$ performs simple harmonic motion with an amplitude of $10 \ cm$ and a time period of $2 \ s$. What is its kinetic energy at a displacement of $5 \ cm$?

Find the displacement of a simple harmonic oscillator at which its $PE$ is half of the maximum energy of the oscillator.

In simple harmonic motion,the total mechanical energy of a given system is $E$. If the mass of the oscillating particle $P$ is doubled,then the new energy of the system for the same amplitude is

In $SHM$,the restoring force is $F = -kx$,where $k$ is the force constant,$x$ is the displacement,and $A$ is the amplitude of motion. Then,the total energy depends upon:

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