Question

A particle is describing simple harmonic motion. If its velocities are $v_1$ and $v_2$ when the displacements from the mean position are $y_1$ and $y_2$ respectively, then its time period is

• A. $2\pi\sqrt{\frac{y^{2}_{1}+y^{2}_{2}}{v^{2}_{1}+v^{2}_{2}}}$
• B. $2\pi\sqrt{\frac{v^{2}_{2}-v^{2}_{1}}{y^{2}_{1}-y^{2}_{2}}}$
• C. $2\pi\sqrt{\frac{v^{2}_{1}-v^{2}_{2}}{y^{2}_{1}+y^{2}_{2}}}$
• D. $2\pi\sqrt{\frac{y^{2}_{1}-y^{2}_{2}}{v^{2}_{2}-v^{2}_{1}}}$

Answer 1

Answer: (D)

In simple harmonic motion,
velocity $v=\omega\sqrt{A^{2}-y^{2}}$
$\therefore \quad v_{1}=\omega\sqrt{A^{2}-y^{2}_{1}}\,\,\Rightarrow \,\,v^{2}_{1}=\omega^{2}A^{2}-\omega^{2}y^{2}_{1}\dots\left(i\right)$
and $\,v_{2}=\omega\sqrt{A^{2}-y^{2}_{2}}\,\,\Rightarrow \,\,v^{2}_{2}=\omega^{2}A^{2}-\omega^{2}y^{2}_{2}\dots\left(ii\right)$
Solving equations (i) and (ii), we get
$v^{2}_{2}-v^{2}_{1}=\omega^{2}\left(y_{1}^{2}-y^{2}_{2}\right)$
$\omega=\sqrt{\frac{v^{2}_{2}-v^{2}_{1}}{y^{2}_{1}-y^{2}_{2}}}$
$\Rightarrow \quad T=\frac{2\pi}{\omega}=2\pi \sqrt{\frac{y^{2}_{1}-y^{2}_{2}}{v^{2}_{2}-v^{2}_{1}}}$