Answer:
When the lift is stationary \[T=2\pi \sqrt{\frac{l}{g}}\] (ii) When the lift accelerates upwards with an acceleration of\[\text{4}.\text{5 m}/{{\text{s}}^{\text{2}}}\], \[T'=2\pi \sqrt{\frac{l}{g+a}}=2\pi \sqrt{\frac{l}{g+4.5}}\] Clearly, the time period decreases when the lift accelerates upwards.
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