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Physics
A body of mass 20 g connected to spring of constant k executes simple harmonic motion with a frequency of ((5/π))Hz. The value of spring constant is
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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
KEAM
KEAM 2007
Oscillations
A
4 N m$^{-1}$
11%
B
3 N m$^{-1}$
0%
C
2 N m$^{-1}$
63%
D
5 N m$^{-1}$
11%
E
2.5 N m$^{-1}$
11%
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 $\, \, \, \, \, \, \, \, \, \, \, \, 10=\sqrt{\frac{k}{0.02}} \, or \, 100 =\frac{k}{0.02}$
$\therefore \, \, \, \, \, \, \, \, \, \, \, \, k=2 Nm^{-1}$