Question

The velocities of a particle executing SHM are $4cm{s}^{-1}$ and $3cm{s}^{-1}$, when its distance from the mean position is $2cm$ and $3cm$ respectively. Calculate its time period.

• A. $5.3s$
• B. $2.94s$
• C. $7.31s$
• D. $9.31s$

Answer 1

Answer: (A)

$5.3s$

Velocity of particle executing SHM at $2cm$ from mean position $v_1=4cm/s$ and at $3cm$ from mean position $v_2=3cm/s$.
$v_1=\omega \sqrt{a^2-x^2}=\omega \sqrt{a^2-4}$

$v_2=\omega \sqrt{a^2-x^2}=\omega \sqrt{a^2-9}$

$\frac{v_2}{v_1}=\frac{\sqrt{a^2-9}}{\sqrt{a^2-4}}=\frac{3}{4}$

$16a^2-144=9a^2-36$

$a^2=\frac{108}{7}$

Putting the value of $a$ in first equation we get :
$\omega =1.18s^{-1}$

Time period, $T=\frac{2\pi }{\omega }=5.3s$