When two sound waves with a phase difference of ` pi //2`, and each having amplitude A and frequency `omega` , are superimposed on each other, then the maximum amplitude and frequency of resultant wave is

Options

A. `(A)/(sqrt(2)),(omega)/(2)`
B. `(A)/(sqrt(2)),omega`
C. `sqrt(2A),(omega)/(2)`
D. `sqrt(2)A,omega`

Answer

D

Explanation

Let `y_(1)=Asin(omegat) and y_(2)=Asin(omegat+(pi)/(2))`
Resultant amplitude
`R^(2)=A^(2)+A^(2)+2A^(2)cos((pi)/(2))`
or `R^(2)=2A^(2)+2A^(2)xx0`
or, `R^(2)=2A^(2)impliesR=sqrt(2)A`
However, both will have the same frequency, on superimposing.