Find the resultant amplitude of the following simple harmonic equations `:`
`x_(1) = 5sin omega t`
`x_(2) = 5 sin (omega t + 53^(@))`
`x_(3) = - 10 cos omega t `

Options

A. 5
B. 10
C. 15
D. 20

Answer

To find the resultant amplitude of the given simple harmonic motions (SHM), we will follow these steps:

### Step 1: Write down the equations of the SHMs
The given SHMs are:
1. \( x_1 = 5 \sin(\omega t) \)
2. \( x_2 = 5 \sin(\omega t + 53^\circ) \)
3. \( x_3 = -10 \cos(\omega t) \)

### Step 2: Convert each SHM into phasor representation
- For \( x_1 = 5 \sin(\omega t) \), this can be represented as a phasor with an amplitude of 5 at an angle of 0°.
- For \( x_2 = 5 \sin(\omega t + 53^\circ) \), this phasor has an amplitude of 5 at an angle of 53°.
- For \( x_3 = -10 \cos(\omega t) \), we convert it to sine form:
\[ x_3 = -10 \cos(\omega t) = -10 \sin\left(\frac{\pi}{2} + \omega t\right) = 10 \sin\left(\omega t + 90^\circ\right) \]
This phasor has an amplitude of 10 at an angle of 90°.

### Step 3: Resolve the phasors into x and y components
1. **For \( x_1 \)**:
- \( x_1 \) components:
- \( x_{1x} = 5 \cos(0^\circ) = 5 \)
- \( x_{1y} = 5 \sin(0^\circ) = 0 \)

2. **For \( x_2 \)**:
- \( x_2 \) components:
- \( x_{2x} = 5 \cos(53^\circ) \)
- \( x_{2y} = 5 \sin(53^\circ) \)

Using \( \cos(53^\circ) \approx \frac{3}{5} \) and \( \sin(53^\circ) \approx \frac{4}{5} \):
- \( x_{2x} = 5 \cdot \frac{3}{5} = 3 \)
- \( x_{2y} = 5 \cdot \frac{4}{5} = 4 \)

3. **For \( x_3 \)**:
- \( x_3 \) components:
- \( x_{3x} = 0 \) (since it is purely vertical)
- \( x_{3y} = -10 \) (since it is directed downwards)

### Step 4: Sum the components
- **Total x-component**:
\[ x_{total} = x_{1x} + x_{2x} + x_{3x} = 5 + 3 + 0 = 8 \]

- **Total y-component**:
\[ y_{total} = x_{1y} + x_{2y} + x_{3y} = 0 + 4 - 10 = -6 \]

### Step 5: Calculate the resultant amplitude
The resultant amplitude \( R \) can be calculated using the Pythagorean theorem:
\[ R = \sqrt{(x_{total})^2 + (y_{total})^2} = \sqrt{(8)^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \]

### Final Result
The resultant amplitude of the given simple harmonic motions is \( \boxed{10} \).