The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is `10 s^(-1)` At, t = 0 the displacement is 5 m. What is the maximum acceleration ? The initial phase is `(pi)/(4)`

Options

A. `500 m//s^(2)`
B. `750 sqrt(2) m//s^(2)`
C. `750 m//s^(2)`
D. `500 sqrt(2) m//s^(2)`

Answer

To solve the problem, we need to find the maximum acceleration in a simple harmonic motion (SHM) given the ratio of maximum acceleration to maximum velocity and the initial conditions.

### Step-by-Step Solution:

1. **Understanding the Relationship**:
In simple harmonic motion, the maximum acceleration \( A_{\text{max}} \) and maximum velocity \( V_{\text{max}} \) are related to the amplitude \( A \) and angular frequency \( \omega \) as follows:
\[ A_{\text{max}} = A \omega^2 \]
\[ V_{\text{max}} = A \omega \]
The ratio of maximum acceleration to maximum velocity is given by:
\[ \frac{A_{\text{max}}}{V_{\text{max}}} = \frac{A \omega^2}{A \omega} = \omega \]

2. **Given Information**:
We know that:
\[ \frac{A_{\text{max}}}{V_{\text{max}}} = 10 \, \text{s}^{-1} \]
Therefore, we can conclude:
\[ \omega = 10 \, \text{s}^{-1} \]

3. **Finding the Amplitude**:
The displacement at \( t = 0 \) is given as 5 m, and the initial phase \( \theta \) is \( \frac{\pi}{4} \). The displacement in SHM can be expressed as:
\[ x(t) = A \cos(\omega t + \theta) \]
At \( t = 0 \):
\[ x(0) = A \cos(\theta) = A \cos\left(\frac{\pi}{4}\right) = A \cdot \frac{1}{\sqrt{2}} \]
Setting this equal to the given displacement:
\[ A \cdot \frac{1}{\sqrt{2}} = 5 \]
Solving for \( A \):
\[ A = 5 \sqrt{2} \, \text{m} \]

4. **Calculating Maximum Acceleration**:
Now we can calculate the maximum acceleration using the formula:
\[ A_{\text{max}} = A \omega^2 \]
Substituting the values of \( A \) and \( \omega \):
\[ A_{\text{max}} = (5 \sqrt{2}) \cdot (10)^2 = 5 \sqrt{2} \cdot 100 = 500 \sqrt{2} \, \text{m/s}^2 \]

5. **Final Answer**:
The maximum acceleration is:
\[ A_{\text{max}} = 500 \sqrt{2} \, \text{m/s}^2 \]