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
a−b
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B
2a−b3
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C
2a23a−b
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D
None of these
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Solution
The correct option is C2a23a−b 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=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 (c) is the correct answer.