Question

At time $t=0$, the displacement of a simple harmonic oscillator from the mean position is $0.2\text{m}$ and the velocity is $0.5\text{m/s}$. What is the amplitude of oscillator if angular frequency is $2\pi \text{rad/s}$.

• A. $0.273\text{m}$
• B. $0.320\text{m}$
• C. $0.214\text{m}$
• D. $0.152\text{m}$

Answer 1

The correct option is C $0.214\text{m}$

The displacement of a simple harmonic oscillator at an instant is given by

$x\left(t\right)=A\sin (\omega t+\varphi )$

where $A$ is the amplitude

Then, the velocity of the oscillator at an instant is given by

$v\left(t\right)=A\omega \cos (\omega t+\varphi )$

At $t=0\text{s},x=0.2\text{m},v=0.5\text{m/s}$

So applying given condition in $x\left(t\right)\&v\left(t\right)$, we get

$0.2=A\sin \varphi$

$0.5=A\omega \cos \varphi$

$\therefore A=\sqrt{(0.2{)}^2+(\frac{0.5}{\omega }{)}^2}$

$=\sqrt{0.04+\frac{0.25}{{\omega }^2}}(\because \omega =2\pi \text{rad/s})$

$A=0.214\text{m}$

Hence, option $\left(\text{C}\right)$ is correct.