MHT CET · Physics · Oscillations

A person measures a time period of a simple pendulum inside a stationary lift and finds it to be T. If the lift starts accelerating upwards with an acceleration $(\frac{g}{3}),$ the time period of the pendulum will be

A $\frac{T}{\sqrt{3}}$
B $\sqrt{3} \frac{T}{2}$
C $\sqrt{3} T$
D $\frac{T}{3}$

Verified Solution

Answer & Solution

Correct Answer

(B) $\sqrt{3} \frac{T}{2}$

Step-by-step Solution

Detailed explanation

Time period of simple pendulum inside a stationary lift,

T = 2 π \sqrt{\frac{l}{g}}

… (i)
Where,

l

= length of string
and g = gravitational acceleration
Acceleration of lift in upward direction,

a = \frac{g}{3}

∴

Time period of pendulum,

T^{'} = 2 π \sqrt{\frac{l}{g + a}} = 2 π \sqrt{\frac{l}{g + \frac{g}{3}}}

= 2 π \sqrt{\frac{3 l}{4 g}} = 2 π . \frac{\sqrt{3}}{2} . \sqrt{\frac{l}{g}}

= \frac{\sqrt{3}}{2} . 2 π \sqrt{\frac{l}{g}} = \frac{\sqrt{3}}{2} T

[from Eq. (i)]

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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Step-by-step Solution

A person measures a time period of a simple pendulum inside a stationary lift and finds it to be T. If the lift starts accelerating upwards with an acceleration $(\frac{g}{3}),$ the time period of the pendulum will be