MHT CET · Physics · 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
- A 2
- B \(\sqrt{2}\)
- C \(2 \sqrt{2}\)
- D \(\frac{1}{\sqrt{2}}\)
Answer & Solution
Correct Answer
(C) \(2 \sqrt{2}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& y=A \cos (\pi t+\pi \phi) \\
& v=A \pi \sin (\pi t+\pi \phi) \\
& \text { At } t=0, \frac{y_0}{A}=\cos \pi \phi \\
& -\frac{v_0}{A}=\sin \pi \phi..
\end{aligned}\)
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 \(\mathrm{v}_0=2 \pi \mathrm{~cm} / \mathrm{s}\)
\(2^2+\frac{(2 \pi)^2}{\pi^2}=A^2 \quad \Rightarrow A=2 \sqrt{2}\)
& 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 \(\mathrm{v}_0=2 \pi \mathrm{~cm} / \mathrm{s}\)
\(2^2+\frac{(2 \pi)^2}{\pi^2}=A^2 \quad \Rightarrow A=2 \sqrt{2}\)
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