Two waves, travelling in the same direction through the same region, have equal frequencies, wavelengths and amplitudes. If the amplitude of each wave is 4 mm and the phase difference between the waves is `90^@`, what is the resultant amplitude ?

Answer

`4sqrt2mm`

Explanation

Phase diffeence `phi=pi/2`
`f and lamda` are same so, `omega` is same
`y_1=rsinwt`
`y_2=rsin(wt+pi/2)`
From the principle of superposition
`y=y_1+y_2`
`=rsinwt+rsin(wt+pi/2)`
`=r[sinwt+sin(wt+pi/2)]`
`=r[2sin{((wt+wt+pi/2)/2)}cos{((wt-wt-pi/w))}]`
`y=2rsin(wt+pi/4)cos(-pi/4)`
`rarr y=sqrt2rsin(wt+(pi/4))`
Resultant amplitude
`=sqrt2 r=4sqrt2mm` (because r=4mm)