Question

The path length of oscillation of simple pendulum of \(1\) m is \(8\) cm. Its maximum velocity is (Take \(g=\pi ^{2}\text{m/s}^{2}\))

• A. \(2\pi\) cm/s
• B. \(4\pi\) cm/s
• C. \(8\pi\) cm/s
• D. \(16\pi\) cm/s

Hint:

The path length is twice the amplitude; find the time period and angular frequency, then use \(v=A\omega\).

Answer 1

The Correct Option is B

Step 1: Direct Formula
$\omega=\sqrt{\dfrac{g}{l}}=\sqrt{\dfrac{\pi^{2}}{1}}=\pi$ rad/s.

Step 2: Amplitude
Amplitude is half the path length, $4$ cm.

Step 3: Result
$v_{max}=\omega A=4\pi$ cm/s, option (B).

Final Answer:
With $\omega=\pi$ rad/s and amplitude 4 cm, the maximum speed is $4\pi$ cm/s. \[ \boxed{\text{(B) }4\pi\text{ cm/s}} \]