The resultant wave formed by the superposition of two waves given by y = a (ω t - kx) and y = a (ω t - kx) has an amplitude of

Correct answer: A. 2 \, a The given equations of the waves are y_1 = a (ω t - kx) and y_2 = a (ω t - kx) . The second wave can be written as y_2 = a (ω t - kx + (π)/(2)). The phase difference between the two waves is Δ φ = (π)/(2). The amplitude of the resultant wave is given by A = a_1^2 + a_2^2 + 2a_1 a_2 (Δ φ) . Substituting a_1 = a , a_2 = a , and Δ φ = (π)/(2): A = a^2 + a^2 + 2a^2 ((π)/(2)) A = 2a^2 = 2 a .

Concepts Of Physics MCQ Edition [Volume 1]

Exam: Concepts Of Physics MCQ Edition [Volume 1]
Subject: Physics
Chapter: Wave Motion and Waves on a String

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Wave Motion and Waves on a String

The resultant wave formed by the superposition of two waves given...

The resultant wave formed by the superposition of two waves given by

y = a \sin(\omega t - kx)

and

y = a \cos(\omega t - kx)

has an amplitude of

Options

A $\sqrt{2}\, a$
B $0$
C $2a$
D $a$

Correct answer

A. $\sqrt{2}\, a$

Step-by-step solution

The given equations of the waves are

y_1 = a \sin(\omega t - kx)

and y_2 = a \cos(\omega t - kx)

.<br><br>The second wave can be written as

y_2 = a \sin \left(\omega t - kx +

\dfrac{\pi}{2}\right).<br><br>The phase difference between the two waves is

\Delta \phi =

\dfrac{\pi}{2}.<br><br>The amplitude of the resultant wave is given by

A =

\sqrt{a_1^2 + a_2^2 + 2a_1 a_2 \cos(\Delta \phi)}.<br><br>Substituting

a_1 = a

,

a_2 = a

, and

\Delta \phi =

\dfrac{\pi}{2}:<br><br>

A =

\sqrt{a^2 + a^2 + 2a^2 \cos\left(\dfrac{\pi}{2}\right)}

A =

\sqrt{2a^2} = \sqrt{2}

a

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