The displacement of a particle performing S.H.M. is given by $Y=A \cos [\pi(t .+\phi)]$. If at…

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
  1. 2
  2. $\sqrt{2}$
  3. $2 \sqrt{2}$
  4. $\frac{1}{\sqrt{2}}$

Solution

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

Asked in: MHT CET 2024 (02 May Shift 2)

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