Q. A body of mass 20 g connected to spring of constant $k$ executes simple harmonic motion with a frequency of $\left( \frac{5}{\pi} \right)$ Hz. The value of spring constant is
A) $4\ Nm^{- 1}$
B) 3 $Nm^{- 1}$
C) 2 $Nm^{- 1}$
D) 5 $Nm^{- 1}$
Explanation
(c) Mass ($m$)=20 g=0.02 kg
Frequency ($f$)=$\frac{5}{\pi}$ Hz
Time period of a loaded spring
$T = 2\pi\sqrt{\frac{m}{k}}$
Frequency $(f) = \frac{1}{2\pi}\ \sqrt{\frac{k}{m}}$
$\frac{5}{\pi} = \frac{1}{2\pi}\ \sqrt{\frac{k}{0.02}}$
or $10 = \sqrt{\frac{k}{0.02}}$
or $100 = \frac{k}{0.02}$
$\therefore$ $k = 2\ Nm^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.