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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_________.
 

Answers (1)

best_answer

\begin{aligned} & x(t)=A \sin (\omega +\phi)\\ &V_{(t)}=\frac{d\left(x_{(t)}\right)}{d t}=A \omega \cos (\omega t+\phi)\\ \end{aligned}

\begin{aligned} &\text { At } t=0, \quad x(t)=2 \mathrm{~cm}=A \sin \left(\omega t+\phi \right)\\ \end{aligned}

\begin{aligned} &V(t)=2 \omega=A \omega \cos (\omega t+\phi)\\ \end{aligned}

\begin{aligned} &\tan (\omega t+\phi)=1\\ \end{aligned}

\begin{aligned} &\text { but } t=0 \end{aligned}

\begin{aligned} \therefore \tan (0+\phi)=1\\ \end{aligned}

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

\begin{aligned} &x_{t}=A \sin (\omega t+\phi)\\ \end{aligned}

\begin{aligned} &2=A \sin \left(0+45^{\circ}\right)\\ \end{aligned}

\begin{aligned} &{A=2 \sqrt{2} \mathrm{~cm}}\\ &\therefore x=2 \end{aligned}

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vishal kumar

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