A particle of mass m= 5g is executing simple harmonic motion with an amplitude `0.3`m and time period `pi//5` second. The maximum value of force acting on the particle is

Options

A. 5 N
B. 4 N
C. `0.5` N
D. `0.15`N

Explanation

To find the maximum force acting on a particle executing simple harmonic motion (SHM), we can follow these steps:

### Step 1: Convert mass to kilograms
The mass \( m \) is given as 5 grams. To convert this to kilograms:
\[ m = \frac{5 \text{ g}}{1000} = 0.005 \text{ kg} \]

### Step 2: Identify the amplitude and time period
The amplitude \( A \) is given as 0.3 m, and the time period \( T \) is given as \( \frac{\pi}{5} \) seconds.

### Step 3: Calculate angular frequency \( \omega \)
The angular frequency \( \omega \) is calculated using the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the value of \( T \):
\[ \omega = \frac{2\pi}{\frac{\pi}{5}} = 2 \cdot 5 = 10 \text{ rad/s} \]

### Step 4: Calculate maximum acceleration \( a_{\text{max}} \)
The maximum acceleration in SHM is given by:
\[ a_{\text{max}} = \omega^2 A \]
Substituting the values of \( \omega \) and \( A \):
\[ a_{\text{max}} = (10)^2 \cdot 0.3 = 100 \cdot 0.3 = 30 \text{ m/s}^2 \]

### Step 5: Calculate maximum force \( F_{\text{max}} \)
The maximum force acting on the particle can be calculated using Newton's second law:
\[ F_{\text{max}} = m \cdot a_{\text{max}} \]
Substituting the values of \( m \) and \( a_{\text{max}} \):
\[ F_{\text{max}} = 0.005 \text{ kg} \cdot 30 \text{ m/s}^2 = 0.15 \text{ N} \]

### Final Answer
The maximum value of the force acting on the particle is:
\[ \boxed{0.15 \text{ N}} \]

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