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 use the relationship between maximum speed, angular frequency, and amplitude.

### Step-by-step Solution:

1. **Identify Given Values**:
- Amplitude (A) = 4 cm = 0.04 m (convert to meters)
- Maximum speed (V_max) = 8 m/s

2. **Relate Maximum Speed to Angular Frequency**:
The maximum speed in SHM is given by the formula:
\[ V_{max} = A \omega \]
where \( \omega \) is the angular frequency.

3. **Solve for Angular Frequency (\(\omega\))**:
Rearranging the formula gives:
\[ \omega = \frac{V_{max}}{A} \]
Substituting the known values:
\[ \omega = \frac{8 \, \text{m/s}}{0.04 \, \text{m}} = 200 \, \text{rad/s} \]

4. **Calculate Maximum Acceleration**:
The maximum acceleration in SHM is given by:
\[ a_{max} = A \omega^2 \]
Substituting the values we have:
\[ a_{max} = 0.04 \, \text{m} \times (200 \, \text{rad/s})^2 \]
\[ a_{max} = 0.04 \, \text{m} \times 40000 \, \text{(rad/s)}^2 = 1600 \, \text{m/s}^2 \]

5. **Final Answer**:
The maximum acceleration of the block is:
\[ a_{max} = 1600 \, \text{m/s}^2 \]