Question

An oscillating pendulum suspended from the roof of a lift which is at rest has time period \(T_1\). When lift moves up with acceleration 'a' its time period is \(T_2\). When lift moves down with acceleration 'a' its time period is \(T_3\). The relation between \(T_1\), \(T_2\) and \(T_3\) is

• A. \(T_1 = \frac{2T_2T_3}{\sqrt{T_2^2+T_3^2}}\)
• B. \(T_1 = \frac{\sqrt{2}T_2T_3}{\sqrt{T_2^2+T_3^2}}\)
• C. \(T_1 = \frac{2T_2^2T_3^2}{\sqrt{T_2+T_3}}\)
• D. \(T_1 = \frac{\sqrt{2}T_2^2T_3^2}{\sqrt{T_2+T_3}}\)

Hint:

Effective g is g + a going up and g - a going down; add 1/T^2 values.

Answer 1

The Correct Option is B

Step 1: Test with a = 0:
If $a=0$, then $T_2=T_3=T_1$. Put this in each option.

Step 2: Check the options:
A: $\dfrac{2T^2}{\sqrt2T}=\sqrt2T\neq T$. B: $\dfrac{\sqrt2T^2}{\sqrt2T}=T$, correct. C: $\dfrac{2T^4}{\sqrt{2T}}$ does not reduce to $T$ dimensionally. D similarly fails.

Step 3: Pick:
Only B passes this test, and it matches the derivation $\dfrac{2}{T_1^2}=\dfrac{1}{T_2^2}+\dfrac{1}{T_3^2}$. Option B.

Final Answer:
Adding 1/T2^2 and 1/T3^2 gives 2/T1^2, which leads to option B. \[ \boxed{\text{(B) }T_1=\dfrac{\sqrt2\,T_2T_3}{\sqrt{T_2^2+T_3^2}}} \]