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: Understanding the Concept
The path length of oscillation is the full distance between the two extremes, \(2A\). So \(A=\dfrac{8}{2}=4\) cm.

Step 2: Angular Frequency
\(T=2\pi\sqrt{\dfrac{l}{g}}=2\pi\sqrt{\dfrac{1}{\pi^{2}}}=2\) s
\(\omega=\dfrac{2\pi}{T}=\pi\) rad/s

Step 3: Maximum Velocity
\(v_{max}=A\omega=4\times\pi=4\pi\) cm/s

Step 4: Compare the options
This matches option (B).

Final Answer:
The maximum velocity is \(4\pi\) cm/s. \[ \boxed{\text{(B) }4\pi\text{ cm/s}} \]