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Q.

Two waves have equations $x_{1}$ = a sin $(ωt + ϕ_{1})$ and $x_{2}$ = a sin $(ωt + ϕ_{2})$ . If in the resultant wave the frequency and amplitude remain equal to amplitude of superimposing waves, the phase difference between them is

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a

$\frac{π}{6}$

b

$\frac{2 π}{3}$

c

$\frac{π}{4}$

d

$\frac{π}{3}$

answer is B.

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Detailed Solution

$\begin{aligned} \sin A + \sin B = 2 \sin (\frac{A + B}{2}) \cos (\frac{A - B}{2}) \\ x_{Net} = 2 a \sin [ω t + (\frac{Φ_{1} + Φ_{2}}{2})] \cos (\frac{Φ_{1} - Φ_{2}}{2}) \\ = 2 a \cos (\frac{Φ_{1} - Φ_{2}}{2}) \sin (ω t + \frac{Φ_{1} + Φ_{2}}{2}) \end{aligned}$

Amplitude of Resultant wave according to the question

$\begin{aligned} 2 acos (\frac{Φ_{1} - ϕ_{2}}{2}) = a \\ \cos (\frac{Φ_{1} - Φ_{2}}{2}) = \frac{1}{2} \\ \Rightarrow \frac{Φ_{1} - Φ_{2}}{2} = \frac{π}{3} \\ Φ_{1} - Φ_{2} = \frac{2 π}{3} \end{aligned}$

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Two waves have equations x1= a sin ωt+ϕ1and x2 = a sin ωt+ϕ2. If in the resultant wave the frequency and amplitude remain equal to amplitude of superimposing waves, the phase difference between them is