Question

A particle suspended from a vertical spring oscillates $10$ times per second. At the highest point of oscillation, the spring becomes unstretched. Find the speed when the spring is stretched by $0.20$cm. (Take, $g={\pi }^{2}m/s^2$).

• A. $110.25cm/s$
• B. $12.14cm/s$
• C. $15.4cm/s$
• D. $16.7cm/s$

Answer 1

The correct option is C $15.4cm/s$

When a spring is in equilibrium its extension $=\frac{mg}{k}$
At the highst point of block, spring is in its natural length.
Hence, amplitude, $A=\frac{mg}{k}$
Given, $f=10$Hz
$\omega =2\pi f=20\pi rad/s$
$\therefore \omega =\sqrt{\frac{k}{m}}$
On squaring both sides, we get
$\frac{k}{m}={\omega }^2$
$\Rightarrow v_{max}=A\omega$
$=\frac{m}{k}g\omega =\frac{g\omega }{{\omega }^2}$
$=\frac{g}{\omega }=\frac{{\pi }^2}{20\pi }=\frac{\pi }{20}m/s$
$\therefore A=\frac{mg}{k}=\frac{g}{{\omega }^2}$
$=\frac{{\pi }^2}{400{\pi }^2}=\frac{0.25}{100}m=0.25$cm
Now, when extension is $0.20$cm, then displacement from equilibrium position$=0.25-0.20=0.05$cm
$x=0.05$cm
$\therefore v=\omega \sqrt{A^2-x^2}$
$=20\pi \sqrt{(0.25{)}^2-(0.05{)}^2}$
$=2\pi \sqrt{6}cm/s$
$=15.4cm/s$.