A particle executing S.H.M. having amplitude 0.01 m and frequency 60 Hz. Determine maximum acceleration of particle.

Options

A. (a)`144 pi^2m//s^2`
B. (b)`120 pi^2m//s^2`
C. (c)`80 pi^2m//s^2`
D. `60 pi^2m//s^2`

Answer

To solve the problem of finding the maximum acceleration of a particle executing simple harmonic motion (SHM) with a given amplitude and frequency, we can follow these steps:

### Step 1: Identify the given values
- Amplitude (A) = 0.01 m
- Frequency (f) = 60 Hz

### Step 2: Calculate the angular frequency (ω)
The angular frequency (ω) is related to the frequency (f) by the formula:
\[ \omega = 2\pi f \]
Substituting the given frequency:
\[ \omega = 2\pi \times 60 = 120\pi \, \text{rad/s} \]

### Step 3: Use the formula for maximum acceleration
The maximum acceleration (a_max) in SHM is given by:
\[ a_{\text{max}} = \omega^2 A \]
Substituting the values of ω and A:
\[ a_{\text{max}} = (120\pi)^2 \times 0.01 \]

### Step 4: Calculate ω²
Calculating \( (120\pi)^2 \):
\[ (120\pi)^2 = 14400\pi^2 \]

### Step 5: Substitute and simplify
Now substituting back into the equation for maximum acceleration:
\[ a_{\text{max}} = 14400\pi^2 \times 0.01 \]
\[ a_{\text{max}} = 144\pi^2 \, \text{m/s}^2 \]

### Step 6: Final result
Thus, the maximum acceleration of the particle is:
\[ a_{\text{max}} = 144\pi^2 \, \text{m/s}^2 \]