JEE Main 26-Aug-2021 Shift 2 Solved Paper

Given, equation of wave of first string, y1=A1‌sin‌k(x−vt)
Equation of wave of second string, y2=A2‌sin‌k(x−vt+x0)
A1=12‌mm,A2=5‌mm,x0=3.5‌cm and k=6.28cm−1
The equation of waves after substitution of values can beexpressed as
y1=12‌sin‌6.28(x−vt) and y2=5‌sin‌6.28(x−vt+3.5)
The phase difference between two waves can be calculated as
Δϕ=

2π

λ

Δx
From equation of waves, the value of
Δx=(x−vt+3.5)−(x−vt)=3.5‌cm
We know that, wave number is k = 2 π / λ.
Now, the phase difference is
∆ϕ = k∆x= 6.28 × 3.5 = 2 π × 3.5 = 7 π
The amplitude of resulting wave can be calculated by using thefollowing relation,
Anet=√A12+A22+2A1A2‌cos‌Δ‌ϕ
Substituting the values in above expression,
Anet=√122+52+2×12×5×cos‌7‌π
=√144+25+120×(−1)
Anet=√49=7‌mm
Thus, the amplitude of resultant wave is 7 mm.