Question:medium
The displacement of a particle executing simple harmonic motion with time period \(T\) is expressed as
\[
x(t)=A\sin\omega t,
\]
where \(A\) is the amplitude of oscillation.
If the maximum value of the potential energy of the oscillator is found at
\[
t=\frac{T}{2\beta},
\]
then the value of \(\beta\) is ________.
The displacement of a particle executing simple harmonic motion with time period \(T\) is expressed as
\[
x(t)=A\sin\omega t,
\]
where \(A\) is the amplitude of oscillation.
If the maximum value of the potential energy of the oscillator is found at
\[
t=\frac{T}{2\beta},
\]
then the value of \(\beta\) is ________.
Show Hint
In SHM, potential energy is maximum at extreme positions and zero at the mean position.
Updated On: Feb 24, 2026
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Correct Answer: 2
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