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, $g=\pi^2 \mathrm{~m} / \mathrm{s}^2$ )

• A. $110.25cms^{- 1}$
• B. $12.14cms^{- 1}$
• C. $15.4cms^{- 1}$
• D. $16.7cms^{- 1}$

Answer 1

Answer: (C)

When a spring is in equilibrium, its extension $=\frac{m g}{k}$ At the highest point of block, spring is in its natural length.
Hence, amplitude, $A=\frac{m g}{k}$
Given, $f=10 \, Hz$
$\omega =2\pi f=20\pi \, rads^{- 1}$
$\therefore \, \omega =\sqrt{\frac{k}{m}}$
On squaring both sides, we get
$\frac{k}{m}=\omega ^{2}$
$\Rightarrow \, v_{m a x}=A\omega $
$=\frac{m}{k}g\omega =\frac{g \omega }{\omega ^{2}}$
$=\frac{g}{\omega }=\frac{\pi ^{2}}{20 \pi }=\frac{\pi }{20}ms^{- 1}$
$\therefore \, \, A=\frac{m g}{k}=\frac{g}{\omega ^{2}}$
$=\frac{\pi ^{2}}{400 \, \pi ^{2}}=\frac{0.25}{100}m=0.25 \, cm$
Now, when the extension is $0.20 \, cm,$ then displacement from the equilibrium position $=0.25-0.20=0.05 \, cm$
$x=0.05 \, cm$
$\therefore \, v=\omega \sqrt{A^{2} - x^{2}}$
$=20\pi \sqrt{\left(0.25\right)^{2} - \left(0.05\right)^{2}}$
$=2\pi \sqrt{6} \, cms^{- 1}$
$=15.4 \, cms^{- 1}$