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Question

A particle executes simple harmonic motion (amplitude = A) between x = -A and x = +A. The time taken for it to go from 0 to A2 is T1 and to go from A2 to A is T2. Then


A

T1<T2

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B

T1>T2

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C

T1=T2

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D

T1=2T2

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Solution

The correct option is A

T1<T2


Using x = A sin ωt

For x = A2 , sin ωT1=12T1=π6ω

For x = A , sin ω(T1 + T2)=1T1+T2=π2ω

T2=π2ωT1=π2ωπ6ω=π3ω i.e T1<T2

Alternate method: In S.H.M., velocity of particle also oscillates simple harmonically.

Speed is more near the mean position and less near the extreme position.

Therefore the time taken for the particle to go from 0 to A2 will be less than

the time taken to go from A2 to A. Hence T1<T2 .


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