Question
A particle is executing a simple harmonic motion. Its maximum acceleration is $$\alpha $$ and maximum velocity is $$\beta .$$ Then, its time period of vibration will be
A.
$$\frac{{{\beta ^2}}}{{{\alpha ^2}}}$$
B.
$$\frac{\alpha }{\beta }$$
C.
$$\frac{{{\beta ^2}}}{\alpha }$$
D.
$$\frac{{2\pi \beta }}{\alpha }$$
Answer :
$$\frac{{2\pi \beta }}{\alpha }$$
Solution :
For a particle executing $$SHM,$$ we have maximum acceleration,
$$\alpha = A{\omega ^2}\,......\left( {\text{i}} \right)$$
where, $$A$$ is maximum amplitude and $$\omega $$ is angular velocity of a particle.
Maximum velocity, $$\beta = A\omega \,......\left( {{\text{ii}}} \right)$$
Dividing Eq. (i) by Eq. (ii), we get
$$\eqalign{
& \frac{\alpha }{\beta } = \frac{{A{\omega ^2}}}{{A\omega }} \cr
& \Rightarrow \frac{\alpha }{\beta } = \omega = \frac{{2\pi }}{T} \cr
& {\text{i}}{\text{.e}}{\text{.}}\,\,T = \frac{{2\pi \beta }}{\alpha } \cr} $$
Thus, its time period of vibration, $$T = \frac{{2\pi \beta }}{\alpha }$$