The acceleration of a certain simple harmonic oscillator is given by `a = -(15.8 m//s^(2)) cos (2.51 t)`. What is the amplitude of the simple harmonic mootion?

Options

A. 2.51m
B. 4/41m
C. 6/30m
D. 11.1m

Answer

To find the amplitude of the simple harmonic motion given the acceleration equation \( a = - (15.8 \, \text{m/s}^2) \cos(2.51 t) \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Form of the Acceleration Equation**:
The acceleration \( a \) of a simple harmonic oscillator can be expressed as:
\[ a = -\omega^2 x \]
where \( \omega \) is the angular frequency and \( x \) is the displacement.

2. **Determine the Maximum Acceleration**:
The maximum value of the acceleration occurs when \( \cos(2.51 t) = 1 \). Thus, the maximum acceleration \( a_{\text{max}} \) can be given by:
\[ a_{\text{max}} = -15.8 \, \text{m/s}^2 \]

3. **Relate Maximum Acceleration to Amplitude**:
From the equation \( a = -\omega^2 x \), we can relate the maximum acceleration to amplitude \( A \):
\[ a_{\text{max}} = \omega^2 A \]
Therefore, we can write:
\[ 15.8 = \omega^2 A \]

4. **Identify the Angular Frequency**:
From the given equation, we see that \( \omega = 2.51 \, \text{rad/s} \).

5. **Substitute \( \omega \) into the Equation**:
Substitute \( \omega \) into the equation for maximum acceleration:
\[ 15.8 = (2.51)^2 A \]

6. **Solve for Amplitude \( A \)**:
Rearranging the equation gives:
\[ A = \frac{15.8}{(2.51)^2} \]
Now, calculate \( (2.51)^2 \):
\[ (2.51)^2 = 6.3001 \]
Then, substitute this value back into the equation for \( A \):
\[ A = \frac{15.8}{6.3001} \approx 2.51 \, \text{m} \]

### Final Answer:
The amplitude of the simple harmonic motion is approximately \( 2.51 \, \text{m} \).