Question

Two waves of amplitude A and xA pass through a region . If x1, the difference in the maximum and minimum resultant amplitude possible is

1. (x +1)A 2.(x-1)A 3.2xA 4. 2A

Answer 1

if two waves of amplitude $A_1$ and $A_2$ and phase differences between them is $\phi$ then, resultant amplitute is given by,

$A_{r}=\sqrt{A_1^2+A_2^2+2A_1A_2cos\phi}$

case 1 : at $\phi=0^{\circ}$ we get, maximum resultant amplitute

so, $A_{max}=|A_1+A_2|$

case 2 : at $\phi=180^{\circ}$ we get, minimum resultant amplitute,

$A_{min}=|A_1-A_2|$

here, we assume, $A_1=xA$ and $A_2=A$

so, $A_{max}=|xA+A|=|(x+1)A|$

and $A_{min}=|xA-A|=|(x-1)A|$

so, difference in the maximum and minimum amplitude possible is |(x + 1)A| - |(x - 1)A| = 2A , if we assume x > 1

and if 0 < x < 1 then, |(x + 1)A| - |(x - 1)A| = 2xA

hence, option (3) and (4) both are possible.

Answer 2

Answer:2Ao

Explanation:Clearly:-

Amax=Ao+xAo .....(1)

Amin =xAo-Ao .....(2)

So the difference between two is :-

Amax -Amin =

(xAo+Ao)-(xAo-Ao)

Which is equal to 2Ao

It was the simplest solution of this question...