Question 295: Oscillations

Question

The displacement of a particle performing S.H.M. is given by $Y=A \cos [\pi(t+\phi)]$. If at $\mathrm{t}=0$, the displacement is $\mathrm{y}=2 \mathrm{~cm}$ and velocity is $2 \pi \mathrm{~cm} / \mathrm{s}$, the value of amplitude $A$ in cm is

Accepted Answer

$$\begin{aligned}
&\begin{aligned}
& y=A \cos (\pi t+\pi \phi)\
& v=A \pi \sin (\pi t+\pi \phi) \
& 	ext { At } t=0, \frac{y_0}{A}=\cos \pi \phi  \quad	ext{.... (i)}\
& -\frac{v_0}{A}=\sin \pi \phi \quad	ext{.... (ii)}
\end{aligned}\
&	ext { Squaring and adding (i) and (ii) }\
&\mathrm{y}_0^2+\frac{\mathrm{v}_0^2}{\pi^2}=\mathrm{A}^2
\end{aligned}$$
$$\begin{aligned}
&	ext { Putting } y_0=2 \mathrm{~cm} 	ext { and } v_0=2 \pi \mathrm{~cm} / \mathrm{s}\
&2^2+\frac{(2 \pi)^2}{\pi^2}=A^2 \quad \Rightarrow A=2 \sqrt{2}
\end{aligned}$$

Answer

2

Answer

$\sqrt{2}$

Answer

$2 \sqrt{2}$

Answer

$\frac{1}{\sqrt{2}}$

Oscillations

The displacement of a particle performing S.H.M. is given by Y=A \cos [\pi(t+\phi)]. If at \mathrm{t}=0, the displacement is \mathrm{y}=2 \mathrm{~cm} and velocity is 2 \pi \mathrm{~cm} / \mathrm{s}, the value of amplitude A in cm is

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