Question

The time taken by a particle executing simple harmonic motion to move from the mean position to $(1/4{)}^{th}$ of the maximum displacement is $1\text{s}$. Calculate the angular frequency of oscillation.
$\left[{\sin }^{-1}\right(1/4)=0.252]$

• A. $0.25\frac{\text{rad}}{s}$
• B. $0.32\frac{\text{rad}}{s}$
• C. $0.15\frac{\text{rad}}{s}$
• D. $0.20\frac{\text{rad}}{s}$

Answer 1

The correct option is A $0.25\frac{\text{rad}}{s}$

The equation of Simple Harmonic Oscillation is given by,

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

Let us consider the particle starts from mean position,

So, $\varphi =0$

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

Given that, $x\left(1\right)=\frac{A}{4}$

$\Rightarrow \frac{A}{4}=A\sin \omega \left(1\right)$

$\Rightarrow \sin \omega =\frac{1}{4}$

$\Rightarrow \omega ={\sin }^{-1}(\frac{1}{4})$

$\Rightarrow \omega =0.252\frac{\text{rad}}{\text{s}}$

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

Why this question ? This question involves basic idea of use of displacement equation to find out other physical quantities.