Question

A simple pendulum performs simple harmonic motion about \(x=0\) with an amplitude 'A' and angular frequency \(\omega\). The speed of pendulum at \(x=\frac{A}{8}\) will be.

• A. \(\frac{\sqrt{61}A\omega }{8}\)
• B. \(\frac{\sqrt{62}A\omega }{8}\)
• C. \(\frac{\sqrt{63}A\omega }{8}\)
• D. \(\frac{\sqrt{65}A\omega }{8}\)

Hint:

Use the speed formula for SHM at displacement x.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
In SHM, \(v=\omega\sqrt{A^{2}-x^{2}}\).

Step 2: Substitute
\(x=\dfrac{A}{8}\), so \(v=\omega\sqrt{A^{2}-\dfrac{A^{2}}{64}}\).

Step 3: Simplify
\(v=\omega A\sqrt{\dfrac{63}{64}}=\dfrac{\sqrt{63}A\omega}{8}\).

Final Answer:
The speed at A/8 is root 63 times A omega divided by 8. \[ \boxed{\text{(C) }\dfrac{\sqrt{63}A\omega}{8}} \]