wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Two waves represented by equation y1=Asin(ωtkx+ϕ1) and y2=Asin(ωtkx+ϕ2) are superimposed such that the amplitude of resultant wave is A. Find the phase difference between them.
Given: (ϕ1>ϕ2)

A
90
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
120
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
60
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
135
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 120
Let phase difference Δϕ=ϕ1ϕ2
We know resultant amplitude is,
Ar=A21+A22+2A1A2cosΔϕ
Given: Ar=A, substituting it in above formula gives
A=A2+A2+2A2cosΔϕ
A=2A2(1+cos Δϕ)
12=1+cos Δϕ
Squaring both sides,
1+cos Δϕ=12
cos Δϕ=12
The negative value of cosΔϕ indicates that it must be an angle of either second or third quadrant.
Δϕ=ππ3=2π3or Δϕ=π+π3=4π3
Δϕ=2π3=120, is the correct answer.

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon