First slide

Applications of SHM

Question

A spring is stretched by 0.2 m when a mass of 0.5 kg is suspended. The time period of the spring when a mass of 0.25kg is suspended and put to oscillation $\frac{\sqrt{2} π}{ψ} s$ . The value of $ψ$ (Given g = 9.8 ms^-2) is ______.

Moderate

Solution

$\begin{array}{l} k x = m g \\ k = \frac{0.5 \times 9.8}{0.2} = \frac{49}{2} {Nm}^{- 1} \\ T = 2 π \sqrt{\frac{m^{'}}{k}} = 2 π \sqrt{\frac{0.25 \times 2}{49}} = 2 π \sqrt{\frac{1}{49 \times 2}} = \frac{\sqrt{2} π}{7} s \\ \Rightarrow ψ = 7 \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App