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Question

# A particle performs SHM in a straight line. In the first second, starting from rest, it travels a distance a and in the next second, it travels a distance b on the same side of the mean position. The amplitude of the SHM is

A
ab
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B
2ab3
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C
2a23ab
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D
None of these
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Solution

## The correct option is C 2a23a−bParticle starts from rest. Hence, we can say that, x=Asin(ωt+π2)=Acos(ωt) .......(1) At t=0,x=A Distance travelled by the particle in one second is given by a=A−Acos(ω×1) ⇒cosω=A−aA=(1−aA) ......(1) Distance travelled by the particle in two seconds is given by a+b=A−Acos(ω×2) ⇒a+b=A−A(2cos2ω−1)=2A−2Acos2ω Using (1) in the above equation, we get a+b=2A−2A(1−aA)2 ⇒a+b=2A(1−(1−aA))(1+1−aA) [∵a2−b2=(a+b)(a−b)] ⇒a+b=2A×aA(2−aA) ⇒a+b=2a[2−aA] ⇒aA=2−a+b2a ⇒A=2a23a−b Thus, option B is the correct answer.

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