Question

A particle executes S.H.M. with time period \(16\) s. At time \(t=2\) s, the particle crosses the mean position while at \(t=4\) s, velocity is \(4\) m/s. The amplitude of motion (in m) is \(\left(\sin 45^{\circ }=\frac{1}{\sqrt{2}}\right)\)

• A. \(\sqrt{2}\pi\)
• B. \(16\sqrt{2}\pi\)
• C. \(24\sqrt{2}\pi\)
• D. \(32\sqrt{2}/\pi\)

Hint:

Take the particle to cross the mean position at t = 2 s, so the displacement is A sin of omega times (t - 2).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
\(\omega=\dfrac{2\pi}{T}=\dfrac{2\pi}{16}=\dfrac{\pi}{8}\) rad/s.
Since the particle is at the mean position at \(t=2\) s, \(y=A\sin\omega(t-2)\).

Step 2: Velocity Expression
\(v=A\omega\cos\omega(t-2)\).

Step 3: At t = 4 s
\(\omega(t-2)=\dfrac{\pi}{8}\times2=\dfrac{\pi}{4}\), so \(v=A\omega\cos45^{\circ}=\dfrac{A\omega}{\sqrt{2}}=4\).

Step 4: Calculation
\(A=\dfrac{4\sqrt{2}}{\omega}=\dfrac{4\sqrt{2}\times8}{\pi}=\dfrac{32\sqrt{2}}{\pi}\) m.

Final Answer:
The amplitude is 32sqrt2/pi metres. \[ \boxed{\text{(D) }32\sqrt{2}/\pi} \]