A simple pendulum has time period '$$T1$$'. The point of suspension is now moved upwards according to the relation $$y=KT2$$.(K=1m/sec2) where y is the vertical displacement. The time period now becomes '$$T2$$'. Then find the ratio of $$T12T22$$ (g=10m/sec2)

Question

A simple pendulum has time period '$$T1$$'. The point of suspension is now moved upwards according to the relation $$y=KT2$$.(K=1m/sec2) where y is the vertical displacement. The time period now becomes '$$T2$$'. Then find  the ratio of  $$T12T22$$  (g=10m/sec2)

Infinity Learn · Accepted Answer

on comparing with equation of motion  for vertical displacement s $$=ut$$ $$+12at2$$, here initial velocity $$u=0$$    $$y=kt2=12at2$$ $$k=12$$×a but given $$k=1$$ $$1=12$$×a acceleration is $$a=2m/sec2$$ time period of simple pendulum is $$T1=2$$$$π$$$$lg---(1)$$ time period of simple pendulum when it is accelerating up $$isT2=2$$$$π$$$$lg+a---(2)$$ divide equation (1) by (2), and subustitute $$a=2$$ $$T12T22=g+ag=10+210=6/5$$

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A simple pendulum has time period 'T1'. The point of suspension is now moved upwards according to the relation y=KT2.(K=1m/sec2) where y is the vertical displacement. The time period now becomes 'T2'. Then find the ratio of T12T22 (g=10m/sec2)

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