| 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\[\frac{\pi }{4}.\] [JEE Online 08-04-2017] |
A) \[500m/{{s}^{2}}\]
B) \[750\sqrt{2}m/{{s}^{2}}\]
C) \[750m/{{s}^{2}}\]
D) \[500\sqrt{2}m/{{s}^{2}}\]
Correct Answer: D
Solution :
| [d] \[{{f}_{\max }}=\omega a\] |
| \[{{V}_{\min }}=a\omega \] |
| \[\frac{\omega a}{a\omega }=10\] |
| W = 10 |
| \[x=a\sin (\omega +\pi /4)\] |
| At f = 0 |
| \[5=a\sin (\pi /4)\] |
| \[a=5\sqrt{2}\] |
| Max acc. = w2a |
| \[=100\times 5\sqrt{2}\] |
| \[=500\sqrt{2}\] |
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