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. $110.25cm{s}^{-1}$
• B. $12.14cm{s}^{-1}$
• C. $15.4cm{s}^{-1}$
• D. $16.7cm{s}^{-1}$

Answer 1

Answer: (C)

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