Question

A wave travels on a light string. The equation of the wave is Y=Asin(kx-wt+30°) . It is reflected from a heavy string tied to an end of the light string at x=0. If 64% of the incident energy is reflected the equation of the reflected wave is
A)Y=0.8Asin(kx-wt+30°+180°) B)Y=0.8Asin(kx+wt+30°+180°)
C)Y=0.8Asin(kx+wt-30°)
D)Y=0.8Asin(kx+wt+30°)

Answer 1

Answer: $Y = 0.8 Asin(kx+\omega t-30\hat{A}^{0})$

Explanation:

Given that,

The equation of the wave is

$Y = Asin(kx-\omega t+30\hat{A}^{0})$

We know, the energy is directly proportional to the square of the amplitude.

$E\propto a^{2}$

$\sqrt E\propto a$.....(I)

In the reflected wave 64% energy is hold on that means,

$E\times 100= 64$

$E = 0.64$

Now, put the value of E in equation (I)

$\sqrt 0.64 \propto a$

$0.8 \propto a$

Now, the reflected wave equation is

$Y = 0.8Asin(kx -\omega t+30\hat{A}^{0})$

But the reflected wave travels in opposite direction

So, the wave equation will be

$Y = 0.8 Asin(kx+\omega t-30\hat{A}^{0})$

Hence, the correct option is c.

Answer 2

"Given:

The equation of the wave is Y=A sin (kx-ωt+300)

We know that  the energy is "directly proportional to the square of the amplitude".

$E\quad \propto \quad { a }^{ 2 }$

$\sqrt{E} \propto a$

Since the reflected wave holds 64% of energy, it can be represented as

E = 0.64

$\sqrt0.64 \propto a$

$0.8 \propto a$

Therefore the reflected wave equation becomes Y=0.8A sin (kx-ωt+300)

Since the reflected wave travels in opposite direction, the equation becomes negative as,

Y= -0.8A sin (kx-ωt+300)"