First slide

Simple harmonic motion

Question

A particle is moving on X-axis and has potential energy $U = 2 - 20 x + 5 x^{2}$ joule, where x is position. The particle is released at x = -3. If the mass of the particle is 0.1 kg, then the maximum velocity (in ms^-1) of the particle is 25 $β$ . If amplitude is 5 m, then the value of $β$ is __________.

Difficult

Solution

$\begin{array}{l} U = 2 - 20 x + 5 x^{2} \\ \frac{d U}{d x} = - 20 + 10 x \\ F = - \frac{d U}{d x} = 20 - 10 x = - 10 (x - 2) \end{array}$
$∴$ The particle is executing SHM about mean position x - 2 = 0 or x = 2
$\begin{array}{l} ∴ k = 10 \\ \Rightarrow m ω^{2} = 10 \\ ω^{2} = \frac{10}{m} = \frac{10}{0.1} = 100 \\ \Rightarrow ω = 10 {rads}^{- 1} \\ v_{max} = A ω = 5 (10) = 50 {ms}^{- 1} \\ ∴ β = 2 \end{array}$

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