MHT CET · Physics · Oscillations
A particle performing SHM has time period \(\frac{2 \pi}{\sqrt{3}}\) and path length \(4 \mathrm{~cm}\). The displacement from mean position at which acceleration is equal to velocity is
- A \(0 \mathrm{~cm}\)
- B \(0.5 \mathrm{~cm}\)
- C \(1 \mathrm{am}\)
- D \(1.5 \mathrm{~cm}\)
Answer & Solution
Correct Answer
(C) \(1 \mathrm{am}\)
Step-by-step Solution
Detailed explanation
Velocity \(v=\omega \sqrt{A^{2}-x^{2}}\)
and acceleration \(=\omega^{2} x\) Given, \(\omega \sqrt{A^{2}-x^{2}}=\omega^{2} x\)
or \(\sqrt{A^{2}-x^{2}}=\omega x\)
Given,
\(
T=\frac{2 \pi}{\sqrt{3}}
\)
and
\(
\omega=\frac{2 \pi}{T}=\sqrt{3}
\)
Substituting the value of \(\omega\) in Eq (i), we get \(\sqrt{A^{2}-x^{2}}=\sqrt{3} x\)
\(
\Rightarrow \quad A=2 x
\)
\(
\begin{aligned}
\text { As amplitude } &=\frac{\text { path length }}{2}=2 \mathrm{~cm} \\
\Rightarrow \quad x=1 \mathrm{~cm}
\end{aligned}
\)
and acceleration \(=\omega^{2} x\) Given, \(\omega \sqrt{A^{2}-x^{2}}=\omega^{2} x\)
or \(\sqrt{A^{2}-x^{2}}=\omega x\)
Given,
\(
T=\frac{2 \pi}{\sqrt{3}}
\)
and
\(
\omega=\frac{2 \pi}{T}=\sqrt{3}
\)
Substituting the value of \(\omega\) in Eq (i), we get \(\sqrt{A^{2}-x^{2}}=\sqrt{3} x\)
\(
\Rightarrow \quad A=2 x
\)
\(
\begin{aligned}
\text { As amplitude } &=\frac{\text { path length }}{2}=2 \mathrm{~cm} \\
\Rightarrow \quad x=1 \mathrm{~cm}
\end{aligned}
\)
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