Question

The stationary wave y = 2a sin kx cos wt in a stretched string is the result of superposition of y1 = a sin (kx – wt) and

• A. y2 = a cos (kx + $\omega$t)
• B. y2 = a sin (kx + $\omega$t)
• C. y2 = a cos (kx – $\omega$t)
• D. y2 = a sin (kx – $\omega$t)

Answer 1

Answer: (B)

y2 = a sin (kx + $\omega$t)

Answer 2

$y_{1} = a\sin(kx - \omega t)$

$y_{2} = a\sin(kx + \omega t)$

According to the principle of superposition, the resultant wave is

$y = y_{1} + y_{2}$

$= a\sin(ks - \omega t) + a\sin(kx + \omega t)$

Using trigonometric identity

$\sin(A + B) + \sin(A - B) = 2\sin A\cos B$

We get $y = 2a\sin kx\cos\omega t$