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=12⇒T1=π6ω For x = A , sin ω(T1 + T2)=1⇒T1+T2=π2ω ⇒T2=π2ω−T1=π2ω−π6ω=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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