For a body in `SHM` the velocity is given by the relation `V = sqrt(144-16x^(2))ms^(-1)`. The maximum acceleration is

Options

A. `12 m//s^(2)`
B. `16 m//s^(2)`
C. `36 m//s^(2)`
D. `48 m//s^(2)`

Explanation

To find the maximum acceleration of a body in simple harmonic motion (SHM) given the velocity equation \( V = \sqrt{144 - 16x^2} \, \text{ms}^{-1} \), we can follow these steps:

### Step 1: Understand the relationship between velocity and displacement in SHM.
In SHM, the velocity \( V \) can be expressed in terms of displacement \( x \) as:
\[ V = \omega \sqrt{A^2 - x^2} \]
where \( \omega \) is the angular frequency and \( A \) is the amplitude.

### Step 2: Compare the given equation with the standard form.
From the given equation:
\[ V = \sqrt{144 - 16x^2} \]
we can rewrite it as:
\[ V = \sqrt{12^2 - (4x)^2} \]
This implies that:
- \( \omega = 12 \, \text{rad/s} \)
- \( A = 3 \, \text{m} \) (since \( 4A = 12 \Rightarrow A = 3 \))

### Step 3: Use the formula for maximum acceleration.
The maximum acceleration \( A_{\text{max}} \) in SHM is given by:
\[ A_{\text{max}} = \omega^2 A \]

### Step 4: Calculate \( A_{\text{max}} \).
Substituting the values of \( \omega \) and \( A \):
\[ A_{\text{max}} = (12)^2 \times 3 = 144 \times 3 = 432 \, \text{m/s}^2 \]

### Conclusion:
The maximum acceleration is \( 432 \, \text{m/s}^2 \).

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