A pendulum suspended from the ceiling of a train has a period T, when the train is at rest. When the train is accelerating with a uniform acceleration a, the period of oscillation will

Increase

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Decrease

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Remain unaffected

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Become infinite

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Solution

The correct option is B
Decrease

Initially time period was T = 2 $π \sqrt{\frac{l}{g}}$
When train accelerates, the effective value of g
Becomes $\sqrt{(g^{2} + a^{2})}$ which is greater than g

Hence, new time period, becomes less than the initial time period.

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A pendulum suspended from the ceiling of a train has a period T, when the train is at rest. When the train is accelerating with a uniform acceleration a, the period of oscillation will

The correct option is B Decrease Initially time period was T = 2 π√lg When train accelerates, the effective value of g Becomes √(g2+a2) which is greater than g Hence, new time period, becomes less than the initial time period.

The correct option is B
Decrease

Initially time period was T = 2 $π \sqrt{\frac{l}{g}}$
When train accelerates, the effective value of g
Becomes $\sqrt{(g^{2} + a^{2})}$ which is greater than g

Hence, new time period, becomes less than the initial time period.