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 side of mean position. The amplitude of the SHM is

Options

A. `(2a^2)/(3b-a)`
B. `(3a^2)/(3a-b)`
C. `(2a^2)/(3a-b)`
D. `(3a^2)/(3b-a)`

Answer

C

Explanation

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)`