First slide

Simple harmonic motion

Question

A particle of mass 10 g is placed in- la- potential field given by $V = (50 x^{2} + 100)$ J/kg. If the frequency of oscillation is found to be $\frac{x}{π}$ cycle/s, the value of x is _______.

Moderate

Solution

Given that potential energy is $U = mV$
$\begin{array}{l} \Rightarrow U = (50 x^{2} + 100) 10^{- 2} \\ F = - \frac{dU}{dx} = - (100 x) 10^{- 2} \\ \Rightarrow {mω}^{2} x = - (100 \times 10^{- 2}) x \\ 10 \times 10^{- 3} ω^{2} x = 100 \times 10^{- 2} x \\ \Rightarrow ω^{2} = 100 \Rightarrow ω = 10 rad / s \\ \Rightarrow f = \frac{ω}{2 π} = \frac{10}{π} = \frac{5}{π} \end{array}$
Hence x = 5.

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