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The time period of simple pendulum is ‘T’ when the length increases by 20cm, its period is T1. When the length is decreased by 20 cm, its time period is T2. When the relation between T,T1 and T2 is

a
2T2=1T12+1T22
b
2T2=1T12−1T22
c
2T2=T12+T22
d
2T2=T12−T22

detailed solution

Correct option is C

time period of a simple pendulum is T=2πlg here 2π, g are constant Tαl on squaring, kT2=l-----(1) kT12=l+20---(2) kT22=l-20---(3) add equation (2)and (3) k(T12+T22)=2l---(4) substitute l from equation (1) k(T12+T22)=2(kT2) T12+T22=2T2

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A simple pendulum has time period T1. The point of suspension is now moved upward according to equation y=kt2  where k=1m/sec2. If new time period is T2 then ratio T12T22  will be


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