Question

A simple harmonic oscillator has an amplitude \( A \) and a time period of \( 6\pi \) seconds. Assuming the oscillation starts from its mean position, the time required by it to travel from \( x = A \) to \( x = \frac{\sqrt{3}}{2} A \) will be \( \frac{\pi}{x} \) seconds, where \( x = \, \underline{\hspace{2cm}} \).

Answer 1

Correct Answer: 2

The provided equation relates the angular frequency \(\omega\) and time \(t\) for a simple harmonic oscillator:

\[ \omega t = \frac{\pi}{6}. \]

The relationship between angular frequency and time period \(T\) is established as:

\[ \omega = \frac{2\pi}{T}. \]

Substituting \(\omega = \frac{2\pi}{T}\) into the initial equation \(\omega t = \frac{\pi}{6}\) yields:

\[ \frac{2\pi}{T} \cdot t = \frac{\pi}{6}. \]

This equation is then simplified to determine \(t\):

\[ t = \frac{\pi}{2} = \frac{\pi}{x}. \]

By comparing the forms \(\frac{\pi}{2} = \frac{\pi}{x}\), the value of \(x\) is determined to be:

\[ x = 2. \]