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 in the same direction. The amplitude of the SHM isA. `(2a^2)/(3b-a)`B. `(3a^2)/(3a-b)`C. `(2a^2)/(3a-b)`D. `(3a^2)/(3b-a)`

A particle performs SHM in a straight line. In the first second, start

1.	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 in the same direction. The amplitude of the SHM isA. `(2a^2)/(3b-a)`B. `(3a^2)/(3a-b)`C. `(2a^2)/(3a-b)`D. `(3a^2)/(3b-a)`
Answer» Correct Answer - C Let the acceleration f, `f=-omega^2x` Therefore distance of the particle from the centre at any time t is given by `x=rcos(omegat)`, where `r` is the amplitude when `t=1s`,`x=r-a` `(r-a)=rcosomega` `cosomega=(r-a)/(r )` .(i) When `t=2`,`x=r-a-b` therefore `r-a-b=cos2omega` `r-a-b=r(2cos^2omega-1)` ..(ii) Substituting the value of `cosomega` from Eq. (i) in Eq. (ii) We get `r-a-b=r[2((r-a)^2)/(r^2)-1]` `=(2(r-a)^2)/(r )-r` `r(3a-b)=2a^2impliesr=(2a^2)/(3a-b)`

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