Question

Two simple pendulums having lengths \(\ell _{1}\) and \(\ell _{2}\) with negligible string mass undergo angular displacement \(\theta _{1}\) and \(\theta _{2}\) from their mean positions respectively. If the angular acceleration of both pendulums are same, then which equation is correct ?

• A. \(\theta _{1}\ell _{2}^{2}=\theta _{2}\ell _{1}^{2}\)
• B. \(\theta _{1}\ell _{1}=\theta _{2}\ell _{2}\)
• C. \(\theta _{1}\ell _{1}^{2}=\theta _{2}\ell _{2}^{2}\)
• D. \(\theta _{1}\ell _{2}=\theta _{2}\ell _{1}\)

Hint:

For small oscillations the angular acceleration is g over l times the angular displacement.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
For small angles a simple pendulum has \(\alpha=-\dfrac{g}{\ell}\theta\).

Step 2: Equal angular acceleration
\(\dfrac{g}{\ell_1}\theta_1=\dfrac{g}{\ell_2}\theta_2\).

Step 3: Simplify
\(\dfrac{\theta_1}{\ell_1}=\dfrac{\theta_2}{\ell_2}\), so \(\theta_1\ell_2=\theta_2\ell_1\).

Step 4: Compare the options
This is option (D).

Final Answer:
The correct relation is theta1 times l2 equals theta2 times l1. \[ \boxed{\text{(D) }\theta_{1}\ell_{2}=\theta_{2}\ell_{1}} \]