Question

The velocities of a particle in $ SHM $ at positions $ X_1 $ and $ X_2 $ are $ V_1 $ and $ V_2 $ respectively. Its time period will be

• A. $ 2\pi\,\sqrt{\left(v^{2}_{1}-v^{2}_{2}\right)/\left(x^{2}_{2}-x^{2}_{1}\right)} $
• B. $ 2\pi\,\sqrt{\left(x^{2}_{1}+x^{2}_{2}\right)/\left(v^{2}_{2}-v^{2}_{1}\right)} $
• C. $ 2\pi\,\sqrt{\left(x^{2}_{2}-x^{2}_{1}\right)/\left(v^{2}_{1}-v^{2}_{2}\right)} $
• D. $ 2\pi\,\sqrt{\left(x^{2}_{2}+x^{2}_{1}\right)/\left(v^{2}_{1}+v^{2}_{2}\right)} $

Answer 1

Answer: (C)

The velocity of SHM
$v = \omega \sqrt{a^{2} -x^{2}}$
According to question
$v_{1} = \omega \sqrt{a^{2} -x_{1}^{2}} \,\,\, ...\left(i\right) $
$v_{2} = \omega\sqrt{a^{2} -x_{2}^{2}} \,\,\, ...\left(ii\right)$
Dividing Eq. $(i)$ by Eq. $(ii)$
$\frac{v_{1}}{v_{2}} = \sqrt{\frac{a^{2}-x_{1}^{2}}{a^{2} - x_{2}^{2}}}$
On both squaring
$\frac{v_{1}^{2}}{v_{2}^{2}} = \frac{a^{2} -x_{1}^{2}}{a^{2}-x_{2}^{2}} $
$a^{2} =\frac{ v_{1}^{2}x_{2}^{2} - v^{2}_{2} x_{1}^{2}}{v_{1}^{2} - v_{2}^{2}} $
Put in Eq. $(i) $
$v_{1} = \omega\sqrt{a^{2}-x_{1}^{2}} $
$v_{1} = \frac{2\pi}{T}\sqrt{\frac{v_{1}^{2}x_{2}^{2} -v_{2}^{2}x_{1}^{2}}{v_{1}^{2} - v_{2}^{2}} -x_{1}^{2}}$
$T = 2\pi\sqrt{\frac{x_{2}^{2} -x_{1}^{2}}{v_{1}^{2} - v_{2}^{2}}} $