Question:
The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is \(10\ \mathrm{s^{-1}}\). At \(t=0\) the displacement is \(5\ \mathrm{m}\). What is the maximum acceleration ? The initial phase is \(\frac{\pi}{4}\).
The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is \(10\ \mathrm{s^{-1}}\). At \(t=0\) the displacement is \(5\ \mathrm{m}\). What is the maximum acceleration ? The initial phase is \(\frac{\pi}{4}\).
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The ratio \(a_{max}/v_{max}\) equals \(\omega\); find the amplitude from \(x(0)=A\sin\varphi\) with \(\varphi=\pi/4\).
Updated On: Nov 11, 2026
- \(500\ \mathrm{m/s^2}\)
- \(500\sqrt2\ \mathrm{m/s^2}\)
- \(750\ \mathrm{m/s^2}\)
- \(750\sqrt2\ \mathrm{m/s^2}\)
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The Correct Option is B
Solution and Explanation
\(a_{max}=\omega^2A,\ v_{max}=\omega A\Rightarrow \omega=\frac{a_{max}}{v_{max}}=10\ \mathrm{s^{-1}}\).
Displacement \(x=A\sin(\omega t+\varphi)\); at \(t=0\): \(5 = A\sin\frac{\pi}{4}=\frac{A}{\sqrt2}\Rightarrow A=5\sqrt2\ \mathrm{m}\).
\(a_{max}=\omega^2A=100\times5\sqrt2=500\sqrt2\ \mathrm{m/s^2}\).
Answer: (2) \(500\sqrt2\ \mathrm{m/s^2}\)
Displacement \(x=A\sin(\omega t+\varphi)\); at \(t=0\): \(5 = A\sin\frac{\pi}{4}=\frac{A}{\sqrt2}\Rightarrow A=5\sqrt2\ \mathrm{m}\).
\(a_{max}=\omega^2A=100\times5\sqrt2=500\sqrt2\ \mathrm{m/s^2}\).
Answer: (2) \(500\sqrt2\ \mathrm{m/s^2}\)
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