MHT CET · Physics · Oscillations
For a particle in SHM, if the amplitude of the displacement is \(a\) and the amplitude of velocity is \(v\), the amplitude of acceleration is
- A \(v a\)
- B \(\frac{v^{2}}{a}\)
- C \(\frac{v^{2}}{2 a}\)
- D \(\frac{v}{a}\)
Answer & Solution
Correct Answer
(B) \(\frac{v^{2}}{a}\)
Step-by-step Solution
Detailed explanation
\(v_{\max }=a \omega\) and
\(
\begin{aligned}
\text { Maximum acceleration } &=\omega^{2} a \\
&=\left(\frac{v}{a}\right)^{2} a=\frac{v^{2}}{a}
\end{aligned}
\)
\(
\begin{aligned}
\text { Maximum acceleration } &=\omega^{2} a \\
&=\left(\frac{v}{a}\right)^{2} a=\frac{v^{2}}{a}
\end{aligned}
\)
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