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
$A=\sqrt{\frac{v_1^2{x}_2^2-v_2^2{x}_1^2}{v_1^2-v_2^2}}$