Question

A particle executes simple harmonic motion represented by displacement function as $x\left(t\right)=A\sin (\omega t+\varphi ).$ If the position and velocity of the particle at $t=0\text{s}$ are $2\text{cm}$ and $2\omega {\text{cm s}}^{-1}$ respectively, then its amplitude is $x\sqrt{2}\text{cm}$ where the value of $x$ is

Answer 1

Particle is executing SHM is represented by
$x\left(t\right)=A\sin (\omega t+\varphi )$
Therefore, we have
$v\left(t\right)=A\omega \cos (\omega t+\varphi )$
Now, since, the position and velocity of the particle at $t=0\text{s}$ are $2\text{cm}$ and $2\omega {\text{cm s}}^{-1}$ respectively
$\Rightarrow 2=A\sin \varphi ...\left(1\right)$
and, $2\omega =A\omega \cos \varphi ...\left(2\right)$
Again, from $\left(1\right)$ and $\left(2\right)$ we have
$\tan \varphi =1$
$\Rightarrow \varphi ={45}^{\circ }$
Putting this value in eq $\left(1\right)$ we get
$2=A\left(\frac{1}{\sqrt{2}}\right)$
$\Rightarrow A=2\sqrt{2}$
Hence, the value of $x=2$.