question_answer

The relation between the acceleration a and displacement \[x\] of a particle executing S.H.M. is\[a=-(p/q)y,\]where p and q are constants. What will be the time period T of the particle?

Answer:

                Here      \[a=-\frac{p}{q}y=-{{\omega }^{2}}y,\]where \[\omega =\sqrt{\frac{p}{q}}\]                 \[\therefore \]  Time period, \[T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{q}{p}}.\]