A body is vibrating in simple harmonic motion with an amplitude of 0.06 m and frequency of 15 Hz . The velocity and acceleration of body is

Options

A. (a)`9.80 ms ^(-1) and 9.03 xx 10^(2) m s^(-2)`
B. (b)`8.90 ms ^(-1) and 8.21 xx10^(2) ms ^(-2)`
C. (c)`6.82 ms ^(-1) and 7.62 xx10^(2) ms ^(-2)`
D. (d)`5.65 ms ^(-1) and 5.32 xx10^(2) ms ^(-2)`

Answer

To solve the problem of a body vibrating in simple harmonic motion (SHM) with an amplitude of 0.06 m and a frequency of 15 Hz, we need to calculate both the velocity and acceleration of the body. Here’s a step-by-step solution:

### Step 1: Calculate Angular Frequency (ω)
The angular frequency (ω) can be calculated using the formula:
\[ \omega = 2\pi f \]
where \( f \) is the frequency.

Given:
- \( f = 15 \, \text{Hz} \)

Substituting the value:
\[ \omega = 2\pi \times 15 = 30\pi \, \text{rad/s} \]

### Step 2: Calculate Velocity (v)
The maximum velocity in SHM can be calculated using the formula:
\[ v = \omega A \]
where \( A \) is the amplitude.

Given:
- \( A = 0.06 \, \text{m} \)

Substituting the values:
\[ v = (30\pi) \times 0.06 \]
Calculating:
\[ v = 30 \times \frac{22}{7} \times 0.06 \]
\[ v = 30 \times \frac{22 \times 6}{700} = \frac{3960}{700} \approx 5.65 \, \text{m/s} \]

### Step 3: Calculate Acceleration (a)
The maximum acceleration in SHM can be calculated using the formula:
\[ a = \omega^2 A \]

Substituting the values:
\[ a = (30\pi)^2 \times 0.06 \]
Calculating:
\[ a = 900\pi^2 \times 0.06 \]
\[ a = 900 \times \left(\frac{22}{7}\right)^2 \times 0.06 \]
\[ a = 900 \times \frac{484}{49} \times 0.06 \]
\[ a = \frac{900 \times 484 \times 6}{49 \times 100} = \frac{2906400}{4900} \approx 593.14 \, \text{m/s}^2 \]

### Final Results
- **Velocity (v)**: \( 5.65 \, \text{m/s} \)
- **Acceleration (a)**: \( 593.14 \, \text{m/s}^2 \)

### Conclusion
Thus, the final answers are:
- Velocity: \( 5.65 \, \text{m/s} \)
- Acceleration: \( 593.14 \, \text{m/s}^2 \)