A simple harmonic motion is represented by `x = 12 sin (10 t + 0.6)`
Find out the maximum acceleration, if displacement is measured in metres and time in seconds.

Options

A. `-1200 m`
B. `-2000 m`
C. `-2400 m`
D. `7200 m`

Explanation

To find the maximum acceleration of the simple harmonic motion represented by the equation \( x = 12 \sin(10t + 0.6) \), we can follow these steps:

### Step 1: Identify the parameters from the equation
The given equation is in the form of:
\[ x = A \sin(\omega t + \phi) \]
where:
- \( A = 12 \) (amplitude)
- \( \omega = 10 \) (angular frequency)
- \( \phi = 0.6 \) (phase constant)

### Step 2: Determine the formula for maximum acceleration
The maximum acceleration \( a_{\text{max}} \) in simple harmonic motion can be calculated using the formula:
\[ a_{\text{max}} = A \omega^2 \]

### Step 3: Substitute the values into the formula
Now, substitute the values of \( A \) and \( \omega \) into the formula:
\[ a_{\text{max}} = 12 \times (10)^2 \]

### Step 4: Calculate the maximum acceleration
Now, calculate the value:
\[ a_{\text{max}} = 12 \times 100 = 1200 \, \text{m/s}^2 \]

### Conclusion
The maximum acceleration of the simple harmonic motion is:
\[ \boxed{1200 \, \text{m/s}^2} \]

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