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. $2\pi \sqrt{\left(v_1^2-v_2^2\right)/\left(x_2^2-x_1^2\right)}$
• B. $2\pi \sqrt{\left(x_1^2+x_2^2\right)/\left(v_2^2-v_1^2\right)}$
• C. $2\pi \sqrt{\left(x_2^2-x_1^2\right)/\left(v_1^2-v_2^2\right)}$
• D. $2\pi \sqrt{\left(x_2^2+x_1^2\right)/\left(v_1^2+v_2^2\right)}$

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. $(i)$
$v_1=\omega \sqrt{a^2-x_1^2}$
$v_1=\frac{2\pi }{T}\sqrt{\frac{v_1^2{x}_2^2-v_2^2{x}_1^2}{v_1^2-v_2^2}-x_1^2}$
$T=2\pi \sqrt{\frac{x_2^2-x_1^2}{v_1^2-v_2^2}}$