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

vertical displacement $$y=ut+12at2$$ here $$u=0$$, hence, $$y=12at2$$ given $$y=kt2$$ comparing, $$y=kt2=12at2$$ ⇒$$k=12$$×a substituting $$k=1$$ ⇒$$1=12$$×a ⇒$$a=2m/sec2$$         $$T1=2$$$$π$$lg         $$T2=2$$$$π$$$$lg+a$$ on dividing above two equations,and substitute $$a=2$$        $$T12T22=g+ag=10+210=6/5$$

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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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