Question

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 ________.

Hint:

In SHM, potential energy is maximum at extreme positions and zero at the mean position.

Answer 1

Correct Answer: 2

Concept: For a particle executing simple harmonic motion (SHM):
Displacement: \( x = A\sin\omega t \)
Angular frequency: \( \omega = \dfrac{2\pi}{T} \)
Potential energy: \[ U = \frac{1}{2}kx^2 \] The potential energy depends on the square of displacement and is {maximum} when the displacement is maximum.
Step 1: Condition for maximum potential energy Maximum potential energy occurs when: \[ |x| = A \] From \( x = A\sin\omega t \): \[ \sin\omega t = \pm 1 \]
Step 2: Find the corresponding time The first time when \( \sin\omega t = 1 \) is: \[ \omega t = \frac{\pi}{2} \Rightarrow t = \frac{\pi}{2\omega} \] Substitute \( \omega = \dfrac{2\pi}{T} \): \[ t = \frac{\pi}{2}\cdot\frac{T}{2\pi}=\frac{T}{4} \]
Step 3: Compare with the given time expression Given: \[ t=\frac{T}{2\beta} \] Equating: \[ \frac{T}{2\beta}=\frac{T}{4} \Rightarrow 2\beta=4 \Rightarrow \beta=2 \] Final Answer: \[ \boxed{2} \]