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

Step 1: Understanding the Concept:
In Simple Harmonic Motion (SHM), the potential energy \(U\) depends on the displacement from the equilibrium position. It reaches its maximum value when the particle is at the extreme positions (i.e., when displacement equals the amplitude).
Step 2: Key Formula or Approach: Potential energy in SHM: \(U = \frac{1}{2} k x^2\).
Maximum potential energy occurs when \(x = \pm A\).
Step 3: Detailed Explanation:
Given displacement: \(x(t) = A \sin \omega t\).
For potential energy to be maximum, \(\sin \omega t\) must be \(\pm 1\).
The first time this occurs (starting from \(t=0\)) is when: \[ \omega t = \frac{\pi}{2} \] We know that \(\omega = \frac{2\pi}{T}\). Substituting this: \[ \left( \frac{2\pi}{T} \right) t = \frac{\pi}{2} \] \[ t = \frac{T}{4} \] The problem states the maximum potential energy is found at \(t = \frac{T}{2\beta}\). Equating the two expressions for time: \[ \frac{T}{4} = \frac{T}{2\beta} \implies 2\beta = 4 \implies \beta = 2 \] Step 4: Final Answer:
The value of \(\beta\) is 2.