A) Tp = T1
B) Tp > Ts
C) Tp < Ts
D) cannot be predicted
Correct Answer: C
Solution :
Time period of simple pendulum placed in a train accelerating at the rate of \[am/{{s}^{2}}\]is given by \[T=2\pi {{\left[ \frac{l}{{{g}^{2}}+{{a}^{2}}} \right]}^{\frac{1}{2}}}\] However, the time period of mass attached to the spring is \[T=2\pi \sqrt{\left( \frac{m}{k} \right)}\] It is independent of g as well as a . Hence, when the train accelerates, the time period of the simple pendulum decreases and that of spring remains unchanged. Hence, \[{{T}_{p}}<T\] and \[{{T}_{s}}=T\] ie, \[{{T}_{p}}<{{T}_{s}}.\]You need to login to perform this action.
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