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)

$2\pi\,\sqrt{\left(x^{2}_{2}-x^{2}_{1}\right)/\left(v^{2}_{1}-v^{2}_{2}\right)}$

Answer 2

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 E $(i)$ by E $(ii)$ $\frac{v_{1}}{v_{2}} = \sqrt{\frac{a^{2}-x_{1}^{2}}{a^{2} - x_{2}^{2}}}$ 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 E $(i)$ $v_{1} = \omega\sqrt{a^{2}-x_{1}^{2}}$ $v_{1} = \frac{2\pi}{T}\sqrt{\frac{v_{1}^{2}x_{2}^{2} -v_{2}^{2}x_{1}^{2}}{v_{1}^{2} - v_{2}^{2}} -x_{1}^{2}}$ $T = 2\pi\sqrt{\frac{x_{2}^{2} -x_{1}^{2}}{v_{1}^{2} - v_{2}^{2}}}$