Question

A particle moves with SHM in a straight liine. In the first second afer starting from rest, it travels a distance `x_(1)` cm and in the next second it travels a distance `x_(2)` cm in the same direction. Prove that the amplitude of oscillation is `2x_(1)^(2)//(3x_(1)-x_(2))`.

Answer 1

As the particle starts from rest, it must start from the extreme position. Hence, (i) when `t=0, x=r` where r is the required amplitude. Using the relation, `x=rcosomegat,` we have
`r=x_(1)=rcosomegaxx1=rcosomega` …(i)
and `r-(x_(1)+x_(2))=rcosomegaxx2=rcos2omega`
or `r-x_(1)-x_(2)=r(2cos^(2)omega-1)` ...(ii)
Solving (i) and (ii) , we get, `r=(2x_(1)^(2))/(3x_(1)-x_(2))`