Two simple harmonic motion are represrnted by the following equation `y_(1) = 40 sin omega t` and `y_(2) = 10 (sin omega t + c cos omega t)`. If their displacement amplitudes are equal, then the value of `c` (in appropriate units) is

Options

A. `sqrt(13)`
B. `sqrt(15)`
C. `sqrt(17)`
D. 4

Explanation

To solve the problem, we need to find the value of \( c \) such that the displacement amplitudes of the two simple harmonic motions represented by the equations \( y_1 = 40 \sin(\omega t) \) and \( y_2 = 10(\sin(\omega t) + c \cos(\omega t)) \) are equal.

### Step-by-Step Solution:

1. **Identify the Amplitude of \( y_1 \)**:
The amplitude of the first simple harmonic motion \( y_1 \) is given directly from the equation:
\[ A_1 = 40 \]

2. **Identify the Amplitude of \( y_2 \)**:
The second equation can be rewritten as:
\[ y_2 = 10 \sin(\omega t) + 10c \cos(\omega t) \]
Here, the amplitude of the sine component is \( 10 \) and the amplitude of the cosine component is \( 10c \).

3. **Calculate the Net Amplitude of \( y_2 \)**:
The net amplitude \( A_2 \) of \( y_2 \) can be calculated using the formula for the resultant amplitude of two perpendicular components:
\[ A_2 = \sqrt{(10)^2 + (10c)^2} \]
Simplifying this gives:
\[ A_2 = \sqrt{100 + 100c^2} = 10\sqrt{1 + c^2} \]

4. **Set the Amplitudes Equal**:
Since the problem states that the amplitudes are equal, we set \( A_1 \) equal to \( A_2 \):
\[ 40 = 10\sqrt{1 + c^2} \]

5. **Solve for \( c \)**:
Dividing both sides by \( 10 \):
\[ 4 = \sqrt{1 + c^2} \]
Now, squaring both sides:
\[ 16 = 1 + c^2 \]
Rearranging gives:
\[ c^2 = 16 - 1 = 15 \]
Taking the square root:
\[ c = \sqrt{15} \]

6. **Conclusion**:
The value of \( c \) is:
\[ c = \sqrt{15} \]