Question

A particle is executing a simple harmonic motion with time period \(T\) and amplitude \(A\) and having the origin at \(O\). What is the time difference between its travel from \(O\) to \(A/2\) and from \(A/2\) to \(A\) on one side of the origin?

• A. \(T/2\)
• B. \(T/4\)
• C. \(T/6\)
• D. \(T/12\)

Hint:

Use \[ x=A\sin\theta \] and convert phase difference into time using \[ t=\frac{T}{2\pi}\theta. \] In SHM, equal distances are generally not covered in equal times.

Answer 1

The Correct Option is D

Step 1: Use the phase-angle picture of SHM.
Starting from the centre, the displacement is $x = A\sin\theta$, where the phase angle $\theta$ grows uniformly with time as $\theta = \frac{2\pi}{T}t$. So equal phase steps take equal time.
Step 2: Phase when the particle reaches $A/2$.
Set $A\sin\theta = \frac{A}{2}$, giving $\sin\theta = \frac{1}{2}$, so $\theta_1 = \frac{\pi}{6}$.
Step 3: Time from O to $A/2$.
\[ t_1 = \frac{T}{2\pi}\cdot\frac{\pi}{6} = \frac{T}{12} \]
Step 4: Phase when the particle reaches $A$.
Set $A\sin\theta = A$, giving $\sin\theta = 1$, so $\theta_2 = \frac{\pi}{2}$.
Step 5: Time from O to $A$.
\[ t_{OA} = \frac{T}{2\pi}\cdot\frac{\pi}{2} = \frac{T}{4} \] So the time from $A/2$ to $A$ is $t_2 = \frac{T}{4} - \frac{T}{12} = \frac{3T - T}{12} = \frac{2T}{12} = \frac{T}{6}$.
Step 6: Take the difference of the two legs.
\[ t_2 - t_1 = \frac{T}{6} - \frac{T}{12} = \frac{2T - T}{12} = \frac{T}{12} \] So the first leg and the gap between the two legs differ by $\frac{T}{12}$.
\[ \boxed{\Delta t = \frac{T}{12}} \]