Question 158: Oscillations

Question

The displacement of a particle, executing simple harmonic motion with time period $T$, is expressed as $x(t)=A \sin \omega t$, where $A$ is the amplitude. The maximum value of potential energy of this oscillator is found at $t=T / 2 \beta$. The value of $\beta$ is $\_\_\_\_$ .

Accepted Answer

The potential energy $(U)$ of a particle executing simple harmonic motion $(S H M)$ is given by :
$$ \mathrm{U}=\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{x}^2 $$
Substituting the displacement equation $\mathrm{x}(\mathrm{t})=\mathrm{A} \sin \omega \mathrm{t}$ :
$$ \mathrm{U}(\mathrm{t})=\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{~A}^2 \sin ^2 \omega \mathrm{t} $$
The potential energy $U$ is maximum when the displacement $x$ is maximum $(x= \pm A)$.
This occurs when :
$$ \sin ^2 \omega t=1 \Rightarrow \sin \omega t= \pm 1 $$
This occurs for a particle starting from the mean position ( $t=0$ ) is at :
$$ \omega \mathrm{t}=\frac{\pi}{2} $$
Using the relationship between angular frequency $\omega$ and time period (T), $\omega=\frac{2 \pi}{\mathrm{~T}}$
$$ \Rightarrow\left(\frac{2 \pi}{T}\right) t=\frac{\pi}{2} \Rightarrow t=\frac{T}{4} $$
It is given that $\mathrm{t}=\frac{\mathrm{T}}{2 \beta}$,
$$ \frac{T}{4}=\frac{T}{2 \beta} \Rightarrow \beta=2 $$
Therefore, the value of $\beta$ is 2 . So, the correct answer is 2 .

Oscillations

The displacement of a particle, executing simple harmonic motion with time period T, is expressed as x(t)=A \sin \omega t, where A is the amplitude. The maximum value of potential energy of this oscillator is found at t=T / 2 \beta. The value of \beta is \_\_\_\_ .

Integer / Numeric Answer Type