A particle executes simple harmonic motion represented by displacement function as

Question:

A particle executes simple harmonic motion represented by displacement function as 

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

If the position and velocity of the particle at $\mathrm{t}=0 \mathrm{~s}$ are $2 \mathrm{~cm}$ and $2 \omega \mathrm{cm} \mathrm{s}^{-1}$ respectively, then its

amplitude is $x \sqrt{2} \mathrm{~cm}$ where the value of $x$ is

Solution:

$\mathrm{x}(\mathrm{t})=\mathrm{A} \sin (\omega \mathrm{t}+\phi)$

$\mathrm{v}(\mathrm{t})=\mathrm{A} \omega \cos (\omega \mathrm{t}+\phi)$

$2=A \sin \phi$           ...(1)

$2 \omega=A \omega \cos \phi$         .....(2)

From (1) and (2)

$\begin{aligned} \tan \phi &=1 \\ \phi &=45^{\circ} \end{aligned}$

Putting value of $\phi$ in equation (1)

$2=A\left\{\frac{1}{\sqrt{2}}\right\}$

$A=2 \sqrt{2}$

$x=2$ 

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