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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 2a23ab
Particle 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=AAcos(ω×1)
cosω=AaA=(1aA) ......(1)
Distance travelled by the particle in two seconds is given by
a+b=AAcos(ω×2)
a+b=AA(2cos2ω1)=2A2Acos2ω
Using (1) in the above equation, we get
a+b=2A2A(1aA)2
a+b=2A(1(1aA))(1+1aA)
[a2b2=(a+b)(ab)]
a+b=2A×aA(2aA)
a+b=2a[2aA]
aA=2a+b2a
A=2a23ab
Thus, option B is the correct answer.


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