JEE Mains · Physics · STD 11 - 13. oscillations
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 \(\frac{\pi }{4}\)
- A \(500\, m/s^2\)
- B \(500\,\sqrt 2 \,m/{s^2}\)
- C \(750\, m/s^2\)
- D \(750\,\sqrt 2 \,m/{s^2}\)
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, \(\frac{A_{\max }}{v_{\max }}=10\)…
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