JEE Mains · Physics · STD 11 - Oscillations
- JEE Mains 2017
- Hard
- MCQ
The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is \(10\,s^{-1}\). At, \(t = 0\) the displacement is \(5\, m\). What is the maximum acceleration ? The initial phase is \(\frac{\pi }{4}\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(500\,\sqrt 2 \,m/{s^2}\)
Step-by-step Solution
Detailed explanation
Maximum velocity in \(\mathrm{SHM}, \mathrm{v}_{\max }=\mathrm{a} \omega\)
Maximum acceleration in \(\mathrm{SHM}, \mathrm{A}_{\max }=\mathrm{a} \omega^{2}\)
where \(a\) and \(\omega\) are maximum amplitude and angular frequency.
Given that, \(\frac{A_{\max }}{v_{\max }}=10\)
i.e., \(\omega=10 \mathrm{s}^{-1}\)
Displacement is given by
Maximum acceleration in \(\mathrm{SHM}, \mathrm{A}_{\max }=\mathrm{a} \omega^{2}\)
where \(a\) and \(\omega\) are maximum amplitude and angular frequency.
Given that, \(\frac{A_{\max }}{v_{\max }}=10\)
i.e., \(\omega=10 \mathrm{s}^{-1}\)
Displacement is given by
\(x=a \sin (\omega t+\pi t 4)\)
at \(t=0, x=5\)
\(5=a \sin \pi / 4\)
\(5=a \sin 45^{\circ} \Rightarrow a=5 \sqrt{2}\)
Maximum acceleration \(\mathrm{A}_{\mathrm{max}}=\mathrm{a} \omega^{2}\)
\(=500\sqrt{2}\,m/s^2\)
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