wave_mechanics interference_and_superposition_of_waves Question 25
Question: The amplitude of a wave represented by displacement equation $ y=\frac{1}{\sqrt{a}}\sin \omega t\pm \frac{1}{\sqrt{b}}\cos \omega t $ will be
[BVP 2003]
Options:
A) $ \frac{a+b}{ab} $
B) $ \frac{\sqrt{a}+\sqrt{b}}{ab} $
C) $ \frac{\sqrt{a}\pm \sqrt{b}}{ab} $
D) $ \sqrt{\frac{a+b}{ab}} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ y=\frac{1}{\sqrt{a}}\sin \omega t\pm \frac{1}{\sqrt{b}}\sin ( \omega t+\frac{\pi }{2} ) $ Here phase difference =$ \frac{\pi }{2} $
$ \therefore $ The resultant amplitude = $ \sqrt{{{( \frac{1}{\sqrt{a}} )}^{2}}+{{( \frac{1}{\sqrt{b}} )}^{2}}}=\sqrt{\frac{1}{a}+\frac{1}{b}}=\sqrt{\frac{a+b}{ab}} $
BETA
