The resultant amplitude due to superposition of two waves
`y_1 =5 sin (wt -kx) and y_2 =-5 cos (wt -kx-150 ^@)`

Options

A. 5
B. `5sqrt(3)`
C. `5sqrt(2-sqrt(3))`
D. `5sqrt(2+sqrt(3))`

Answer

To find the resultant amplitude due to the superposition of the two waves given by \( y_1 = 5 \sin(\omega t - kx) \) and \( y_2 = -5 \cos(\omega t - kx - 150^\circ) \), we can follow these steps:

### Step 1: Rewrite \( y_2 \) in terms of sine
We start with the second wave:
\[ y_2 = -5 \cos(\omega t - kx - 150^\circ) \]
Using the identity \( \cos(\theta) = \sin(90^\circ - \theta) \), we rewrite \( y_2 \):
\[ y_2 = -5 \sin\left(90^\circ - (\omega t - kx - 150^\circ)\right) \]
This simplifies to:
\[ y_2 = -5 \sin\left(240^\circ - \omega t + kx\right) \]
Since \( 240^\circ = 180^\circ + 60^\circ \), we can use the sine identity \( \sin(180^\circ + \theta) = -\sin(\theta) \):
\[ y_2 = 5 \sin(\omega t - kx + 60^\circ) \]

### Step 2: Combine the two waves
Now we can express both waves in terms of sine:
\[ y_1 = 5 \sin(\omega t - kx) \]
\[ y_2 = 5 \sin(\omega t - kx + 60^\circ) \]
We can now add these two waves:
\[ y = y_1 + y_2 = 5 \sin(\omega t - kx) + 5 \sin(\omega t - kx + 60^\circ) \]

### Step 3: Use the sine addition formula
Using the sine addition formula:
\[ \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]
Let \( A = \omega t - kx \) and \( B = \omega t - kx + 60^\circ \):
\[ y = 5 \left( \sin A + \sin B \right) = 5 \left( 2 \sin\left(\frac{(A + B)}{2}\right) \cos\left(\frac{(A - B)}{2}\right) \right) \]
Calculating \( A + B \) and \( A - B \):
\[ A + B = 2(\omega t - kx) + 60^\circ \]
\[ A - B = -60^\circ \]
Thus, we can write:
\[ y = 10 \sin\left(\omega t - kx + 30^\circ\right) \cos\left(-30^\circ\right) \]
Since \( \cos(-30^\circ) = \cos(30^\circ) = \frac{\sqrt{3}}{2} \):
\[ y = 10 \sin\left(\omega t - kx + 30^\circ\right) \cdot \frac{\sqrt{3}}{2} \]
This gives us:
\[ y = 5\sqrt{3} \sin\left(\omega t - kx + 30^\circ\right) \]

### Step 4: Resultant Amplitude
The resultant amplitude \( A \) is:
\[ A = 5\sqrt{3} \]

### Final Result
Thus, the resultant amplitude due to the superposition of the two waves is:
\[ \boxed{5\sqrt{3}} \]