Question

The displacement time (x-t) graph of a particle performing simple harmonic motion is shown in the figure. The acceleration of the particle at \(t=1\) s is

• A. \(\frac{-\pi ^{2}}{4}\) m/s\({}^{2}\)
• B. \(\frac{-\pi ^{2}}{2}\) m/s\({}^{2}\)
• C. \(\frac{\pi ^{2}}{4}\) m/s\({}^{2}\)
• D. \(\frac{\pi ^{2}}{2}\) m/s\({}^{2}\)

Hint:

Read amplitude and period from the graph, then use a = -omega squared x at the given time.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
In SHM the acceleration is \(a=-\omega^2x\), with \(\omega=\dfrac{2\pi}{T}\).

Step 2: Read the graph
The graph gives amplitude \(A=2\text{ m}\) and period \(T=4\text{ s}\), so \(\omega=\dfrac{2\pi}{4}=\dfrac{\pi}{2}\text{ rad/s}\).

Step 3: Position at t = 1 s
At \(t=1\text{ s}\) the particle is at the positive extreme, \(x=+2\text{ m}\).

Step 4: Acceleration
\(a=-\left(\dfrac{\pi}{2}\right)^2\times2=-\dfrac{\pi^2}{2}\text{ m/s}^2\)

Final Answer:
The acceleration is \(-\dfrac{\pi^2}{2}\text{ m/s}^2\). \[ \boxed{\text{(B) }\dfrac{-\pi^{2}}{2}\text{ m/s}^{2}} \]