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Q. A body of mass $20\, g$ connected to spring of constant $k$ executes simple harmonic motion with a frequency of $\bigg(\frac{5}{\pi}\bigg)Hz$. The value of spring constant is

KEAMKEAM 2007Oscillations

Solution:

Mass (m) = 20g = 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 $\, \, \, \, \, \, \, \, \, \, \, \, 10=\sqrt{\frac{k}{0.02}} \, or \, 100 =\frac{k}{0.02}$
$\therefore \, \, \, \, \, \, \, \, \, \, \, \, k=2 Nm^{-1}$