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 given relationship involves the angular frequency \(\omega\) and time \(t\) of a simple harmonic oscillator:

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

We know the relationship between angular frequency and time period \(T\):

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

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

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

Simplify the equation to solve for \(t\):

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

Comparing \(\frac{\pi}{2} = \frac{\pi}{x}\), we find:

\[ x = 2. \]

Answer 2

1) Reference-circle relation.
For SHM, the projection angle made by the radius vector after time \(t\) is \[ \theta=\omega t . \] From the figure, the displacement equals \(A\frac{\sqrt{3}}{2}\). Using the reference circle, \[ \frac{y}{A}=\cos\theta=\frac{\sqrt{3}}{2}\;\;\Longrightarrow\;\;\theta=\frac{\pi}{6}. \] Hence \[ \omega t=\frac{\pi}{6}. \]

2) Express \(\omega\) via the period.
\[ \omega=\frac{2\pi}{T}\quad\Longrightarrow\quad \frac{2\pi}{T}\,t=\frac{\pi}{6} \;\;\Longrightarrow\;\; t=\frac{T}{12}. \]

3) Read the time from the figure.
The marked instant corresponds to a quarter of a period counted from the mean-position crossing (the phasor has swept \(90^\circ\)), so \[ t=\frac{T}{4}=\frac{\pi}{2\omega}. \] Equating with \(t=\dfrac{T}{12}\) from step (2) gives \[ \frac{T}{12}=\frac{\pi}{2\omega}\;\;\Longrightarrow\;\; \omega=\frac{3\pi}{T}. \] Now substitute \(t=\dfrac{\pi}{x}\) into \(\omega t=\dfrac{\pi}{6}\): \[ \omega\left(\frac{\pi}{x}\right)=\frac{\pi}{6} \;\;\Longrightarrow\;\; \frac{\pi}{x}=\frac{1}{2\omega}. \] Using \(\dfrac{1}{2\omega}=\dfrac{\pi}{2}\) from the previous line, \[ \frac{\pi}{x}=\frac{\pi}{2}\;\;\Longrightarrow\;\; x=2. \]

Answer: \(x=2\).