Question

A particle executing simple harmonic motion is having velocities $v_{1}$ and $v_{2}$ at distance $x_{1}$ and $x_{2}$ respectively from the equilibrium position. The amplitude of the motion is

• A. $\sqrt{\frac{v_{1}^{2} x_{2}^{2}-v_{2}^{2} x_{1}^{2}}{v_{1}^{2}+v_{2}^{2}}}$
• B. $\sqrt{\frac{v_{1}^{2} x_{1}^{2}-v_{2}^{2} x_{2}^{2}}{v_{1}^{2}+v_{2}^{2}}}$
• C. $\sqrt{\frac{v_{1}^{2} x_{2}^{2}-v_{2}^{2} x_{1}^{2}}{v_{1}^{2}-v_{2}^{2}}}$
• D. $\sqrt{\frac{v_{1}^{2} x_{2}^{2}+v_{2}^{2} x_{1}^{2}}{v_{1}^{2}+v_{2}^{2}}}$

Answer 1

Answer: (C)

As $v=\omega \sqrt{A^{2}-x^{2}}$
$\therefore v_{1}=\omega \sqrt{A^{2}-x_{1}^{2}}$ or $v_{1}^{2}=\omega^{2}\left(A^{2}-x_{1}^{2}\right) \,\,\,\,\,\,\,\,\,\,\dots(i)$
and $v_{2}=\omega \sqrt{A^{2}-x_{2}^{2}}$ or $v_{2}^{2}=\omega^{2}\left(A^{2}-x_{2}^{2}\right)\,\,\,\,\,\,\,\,\,\,\dots(ii)$
Dividing (i) by (ii), we get
$\frac{v_{1}^{2}}{v_{2}^{2}}=\frac{\left(A^{2}-x_{1}^{2}\right)}{\left(A^{2}-x_{2}^{2}\right)}$
or $v_{1}^{2} A^{2}-v_{1}^{2} x_{2}^{2}=v_{2}^{2} A^{2}-v_{2}^{2} x_{1}^{2}$
or $A^{2}\left(v_{1}^{2}-v_{2}^{2}\right)=v_{1}^{2} x_{2}^{2}-v_{2}^{2} x_{1}^{2}$ or
$A=\sqrt{\frac{v_{1}^{2} x_{2}^{2}-v_{2}^{2} x_{1}^{2}}{v_{1}^{2}-v_{2}^{2}}}$