### Neet Solved Paper 2015 Question 23

##### Question: When two displacements represented by $ y _1=a\sin (\omega t) $ and $ y _2=b\cos ,(\omega t) $ are superimposed, the motion is

#### Options:

A) not a simple harmonic

B) simple harmonic with amplitude $ \frac{a}{b} $

C) simple harmonic with amplitude $ \sqrt{a^{2}+b^{2}} $

D) simple harmonic with amplitude $ \frac{(a+b)}{2} $

## Show Answer

#### Answer:

Correct Answer: C

#### Solution:

Given, $ y _1=a\sin \omega t $ $ y _2=b,\cos \omega t=b\sin ( \omega t+\frac{\pi }{2} ) $

The resultant displacement is given by $ y=y _1+y _2=\sqrt{a^{2}+b^{2}}\sin (\omega t+\phi ) $

Hence, the motion of superimposed wave is simple harmonic with amplitude $ \sqrt{a^{2}+b^{2}}. $