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 step by step, we will use the concepts of simple harmonic motion (SHM).

### Step 1: Understand the given information
We are given:
- The ratio of maximum acceleration (A_max) to maximum velocity (V_max) is \(10 \, \text{s}^{-1}\).
- At \(t = 0\), the displacement \(x = 5 \, \text{m}\).
- The initial phase \(\phi = \frac{\pi}{4}\).

### Step 2: Write the equation of motion for SHM
The general equation for SHM with an initial phase is:
\[ x(t) = A \sin(\omega t + \phi) \]
Substituting the initial phase:
\[ x(t) = A \sin(\omega t + \frac{\pi}{4}) \]

### Step 3: Find the amplitude \(A\)
At \(t = 0\):
\[ x(0) = A \sin\left(\frac{\pi}{4}\right) = 5 \]
Since \(\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\):
\[ A \cdot \frac{1}{\sqrt{2}} = 5 \]
Thus:
\[ A = 5\sqrt{2} \, \text{m} \]

### Step 4: Relate maximum acceleration and maximum velocity
The maximum acceleration in SHM is given by:
\[ A_{\text{max}} = \omega^2 A \]
The maximum velocity in SHM is given by:
\[ V_{\text{max}} = \omega A \]

### Step 5: Set up the ratio of maximum acceleration to maximum velocity
From the problem, we have:
\[ \frac{A_{\text{max}}}{V_{\text{max}}} = 10 \, \text{s}^{-1} \]
Substituting the expressions for maximum acceleration and maximum velocity:
\[ \frac{\omega^2 A}{\omega A} = 10 \]
This simplifies to:
\[ \omega = 10 \, \text{s}^{-1} \]

### Step 6: Calculate the maximum acceleration
Now, substituting \(\omega\) and \(A\) into the formula for maximum acceleration:
\[ A_{\text{max}} = \omega^2 A = (10)^2 (5\sqrt{2}) = 100 \cdot 5\sqrt{2} = 500\sqrt{2} \, \text{m/s}^2 \]

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

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