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 Amplitudes**:
- From the first equation \( y_1 = 40 \sin(\omega t) \), the amplitude \( A_1 \) is \( 40 \).
- From the second equation \( y_2 = 10 (\sin(\omega t) + c \cos(\omega t)) \), we can rewrite it as:
\[ y_2 = 10 \sin(\omega t) + 10c \cos(\omega t) \]
- Here, the amplitude \( A_2 \) can be found using the formula for the resultant amplitude of two perpendicular components:
\[ A_2 = \sqrt{(10)^2 + (10c)^2} = \sqrt{100 + 100c^2} = 10\sqrt{1 + c^2} \]

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

3. **Solve for \( c \)**:
- Divide both sides by 10:
\[ 4 = \sqrt{1 + c^2} \]
- Square both sides to eliminate the square root:
\[ 16 = 1 + c^2 \]
- Rearranging gives:
\[ c^2 = 16 - 1 = 15 \]
- Taking the square root of both sides:
\[ c = \sqrt{15} \]

4. **Final Result**:
- The value of \( c \) is \( \sqrt{15} \).