### 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} $