### Oscillations Question 6

#### Question 6 - 30 January - Shift 1

The general displacement of a simple harmonic oscillator is $x=A \sin \omega t$. Let $T$ be its time period. The slope of its potential energy $(U)$ - time $(t)$ curve will be maximum when $t=\frac{T}{\beta}$. The value of $\beta$ is

## Show Answer

#### Answer: (8)

#### Solution:

#### Formula: Equation of Simple harmonic motion, Potential energy of Simple harmonic motion

$x=A \sin (\omega t)$

$U _{(x)}^{math}=\frac{1}{2} kx^{2}$,

$\frac{dU}{dt}=\frac{1}{2} k 2 x \frac{dx}{dt}$

$=kA^{2} \omega \sin \omega t \cos \omega t \times \frac{2}{2}$

$(\frac{dU}{dt}) _{\max }=\frac{kA^{2} \omega}{2}(\sin 2 \omega t) _{\max }$

$2 \omega t=\frac{\pi}{2} \Rightarrow t=\frac{\pi}{4} \omega=\frac{T}{8} \Rightarrow \beta=8$