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: Understanding the Concept:
The time period of a simple pendulum is \(T=2\pi\sqrt{\dfrac{L}{g_{eff}}}\). In a lift, the effective gravity changes.

Step 2: Write the three periods:
At rest: \(T_1=2\pi\sqrt{\dfrac Lg}\). Lift going up with acceleration \(a\): \(T_2=2\pi\sqrt{\dfrac{L}{g+a}}\). Lift going down: \(T_3=2\pi\sqrt{\dfrac{L}{g-a}}\).

Step 3: Square and invert:
\(\dfrac{1}{T_2^2}=\dfrac{g+a}{4\pi^2L}\) and \(\dfrac{1}{T_3^2}=\dfrac{g-a}{4\pi^2L}\). Add: \(\dfrac1{T_2^2}+\dfrac1{T_3^2}=\dfrac{2g}{4\pi^2L}=\dfrac{2}{T_1^2}\).

Step 4: Solve for T1:
\(T_1^2=\dfrac{2T_2^2T_3^2}{T_2^2+T_3^2}\), so \(T_1=\dfrac{\sqrt2\,T_2T_3}{\sqrt{T_2^2+T_3^2}}\). Option B.

Step 5: Why the other options are wrong.
Option A has 2 instead of \(\sqrt2\). Options C and D have the wrong powers of \(T_2\) and \(T_3\) and do not match the derived form.

Final Answer:
T1 = sqrt 2 T2 T3 / sqrt(T2^2 + T3^2). \[ \boxed{\text{(B) }T_1=\dfrac{\sqrt2\,T_2T_3}{\sqrt{T_2^2+T_3^2}}} \]