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A body of mass 20 g connected to spring of constant k executes simple harmonic motion with a frequency of $(\frac{5}{π}) Hz$ . The value of spring constant (in SI unit) is

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answer is 2.

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Detailed Solution

$\begin{array}{r} \end{array}$

$Mass (m) = 20 g = 0 .02 kg$

$Frequency (f) = \frac{5}{π} Hz$

$Time period of a loaded spring T = 2 π \sqrt{\frac{m}{k}}$

$Frequency (f) = \frac{1}{2 π} \sqrt{\frac{k}{m}}$

$\frac{5}{π} = \frac{1}{2 π} \sqrt{\frac{k}{0 .02}} Or 10 = \sqrt{\frac{k}{0 .02}} Or 100 = \frac{k}{0 .02}$

$∴ k = 2 {Nm}^{- 1}$

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