JEE Main On 8 April 2017 Question 25
Question: 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]
Options:
A) $ 500m/s^{2} $
B) $ 750\sqrt{2}m/s^{2} $
C) $ 750m/s^{2} $
D) $ 500\sqrt{2}m/s^{2} $
Show Answer
Answer:
Correct Answer: D
Solution:
- $ {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} $
