A block of mass m kg hanging from a verticla spring executes simple harmonic motion of amplitude 4 cm . If maximum speed of particle is `8 m//s` . Maximum acceleration of block is

Options

A. `800 m//s^(2)`
B. `100 m//s^(2)`
C. `1600 m//s^(2)`
D. ` 400 m//s^(2)`

Answer

To find the maximum acceleration of a block executing simple harmonic motion (SHM), we can follow these steps:

### Step 1: Understand the given data
- Amplitude (A) = 4 cm = 0.04 m
- Maximum speed (V_max) = 8 m/s

### Step 2: Relate maximum speed to angular frequency
The formula for maximum speed in SHM is given by:
\[ V_{max} = A \cdot \omega \]
Where:
- \( V_{max} \) is the maximum speed,
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency.

### Step 3: Rearrange to find angular frequency
Rearranging the formula to solve for \( \omega \):
\[ \omega = \frac{V_{max}}{A} \]

### Step 4: Substitute the known values
Substituting the known values into the equation:
\[ \omega = \frac{8 \, \text{m/s}}{0.04 \, \text{m}} \]
\[ \omega = 200 \, \text{rad/s} \]

### Step 5: Calculate maximum acceleration
The formula for maximum acceleration in SHM is given by:
\[ a_{max} = A \cdot \omega^2 \]

### Step 6: Substitute the values for amplitude and angular frequency
Substituting the values we have:
\[ a_{max} = 0.04 \, \text{m} \cdot (200 \, \text{rad/s})^2 \]

### Step 7: Calculate \( \omega^2 \)
Calculating \( \omega^2 \):
\[ \omega^2 = 200^2 = 40000 \, \text{rad}^2/\text{s}^2 \]

### Step 8: Final calculation for maximum acceleration
Now substituting \( \omega^2 \) back into the equation for maximum acceleration:
\[ a_{max} = 0.04 \cdot 40000 \]
\[ a_{max} = 1600 \, \text{m/s}^2 \]

### Conclusion
The maximum acceleration of the block is \( 1600 \, \text{m/s}^2 \).

---