Question

A simple pendulum has time period ' \(\mathrm{T}_1\) '. The point of suspension is now moved upward according to equation \(\mathrm{y}=\mathrm{kt}^2\) where \(\mathrm{k}=1 \mathrm{~m} / \mathrm{s}^2\). If new time period is ' \(\mathrm{T}_2\) ' then \(\mathrm{T}_1^2 / \mathrm{T}_2^2\) will be ( \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\) )

• A. \(2/3\)
• B. \(5/6\)
• C. \(6/5\)
• D. \(3/2\)

Hint:

Upward acceleration of the support increases the effective gravity.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The support moves as \(y=kt^{2}\), so its upward acceleration is \(a=2k=2\text{ m/s}^{2}\).

Step 2: Effective gravity
\(g_{eff}=g+a=10+2=12\text{ m/s}^{2}\)

Step 3: Calculation
\(T=2\pi\sqrt{\dfrac{l}{g}}\Rightarrow\dfrac{T_1^{2}}{T_2^{2}}=\dfrac{g_{eff}}{g}=\dfrac{12}{10}=\dfrac{6}{5}\), option (C).

Final Answer:
The ratio T1 squared to T2 squared is 6/5. \[ \boxed{\text{(C) }6/5} \]