Let a wave y(x, t)=a sin (k x-ω t) is reflected from an open boundary and then the incident and reflected waves overlap. Then the amplitude of resultant wave is

Question

Let a wave y(x, t)=a sin (k x-ω t) is reflected from an open boundary and then the incident and reflected waves overlap. Then the amplitude of resultant wave is (A) 2 a cos k x (B) 2 a sin k x (C) 2 a sin ((k x/2)) (D) a sin k x

Tardigrade Admin · Accepted Answer

We have, incident wave, y1=a sin (k x-ω t) So the reflected wave, y2=a sin (k x+ω t) From principle of superposition, The standing wave equation is obtained as y(x, t) =y1+y2=a[ sin (k x-ω t)+ sin (k x+ω t)] =2 a sin k x cos ω t On comparing Eq. (i) with y(x, t)=A(x) cos ω t, we get Amplitude, A(x)=2 a sin k x

Q. Let a wave $y (x, t) = a sin (k x - ω t)$ is reflected from an open boundary and then the incident and reflected waves overlap. Then the amplitude of resultant wave is

$2 a cos k x$

$2 a sin k x$

$2 a sin (\frac{k x}{2})$

$a sin k x$

Q. Let a wave y(x,t)=asin(kx−ωt) is reflected from an open boundary and then the incident and reflected waves overlap. Then the amplitude of resultant wave is

2acoskx

2asinkx

2asin(2kx​)

asinkx

Q. Let a wave $y (x, t) = a sin (k x - ω t)$ is reflected from an open boundary and then the incident and reflected waves overlap. Then the amplitude of resultant wave is

$2 a cos k x$

$2 a sin k x$

$2 a sin (\frac{k x}{2})$

$a sin k x$