A particle executes SHM with amplitude of 20 cm and time period of 12 s. What is the minimum time required for it to move between two points 10 cm on either side of the mean position?

Options

A. 1 s
B. 2 s
C. 3 s
D. 4 s

Explanation

To solve the problem step by step, we will use the principles of Simple Harmonic Motion (SHM).

### Step 1: Understand the parameters given
- Amplitude (A) = 20 cm
- Time period (T) = 12 s

### Step 2: Identify the positions
The particle is moving between two points that are 10 cm on either side of the mean position. Therefore, the two points are:
- Position 1: -10 cm
- Position 2: +10 cm

### Step 3: Write the equation of SHM
The displacement \( x \) in SHM can be described by the equation:
\[ x = A \sin(\omega t) \]
Where:
- \( A \) is the amplitude
- \( \omega \) is the angular frequency

### Step 4: Calculate the angular frequency
The angular frequency \( \omega \) is given by:
\[ \omega = \frac{2\pi}{T} \]
Substituting the value of \( T \):
\[ \omega = \frac{2\pi}{12} = \frac{\pi}{6} \, \text{rad/s} \]

### Step 5: Set up the equation for the position
We need to find the time taken to move from -10 cm to +10 cm. Using the equation of SHM:
For position -10 cm:
\[ -10 = 20 \sin(\omega t_1) \]
\[ \sin(\omega t_1) = -\frac{1}{2} \]

For position +10 cm:
\[ 10 = 20 \sin(\omega t_2) \]
\[ \sin(\omega t_2) = \frac{1}{2} \]

### Step 6: Solve for \( \omega t_1 \) and \( \omega t_2 \)
From \( \sin(\omega t_1) = -\frac{1}{2} \):
\[ \omega t_1 = \frac{7\pi}{6} \, \text{(1st quadrant)} \text{ or } \frac{11\pi}{6} \, \text{(4th quadrant)} \]

From \( \sin(\omega t_2) = \frac{1}{2} \):
\[ \omega t_2 = \frac{\pi}{6} \, \text{(1st quadrant)} \text{ or } \frac{5\pi}{6} \, \text{(2nd quadrant)} \]

### Step 7: Calculate the time intervals
Using \( \omega = \frac{\pi}{6} \):
- For \( t_1 \):
\[ t_1 = \frac{7\pi/6}{\pi/6} = 7 \, \text{s} \]

- For \( t_2 \):
\[ t_2 = \frac{\pi/6}{\pi/6} = 1 \, \text{s} \]

### Step 8: Find the total time taken
The total time taken to move from -10 cm to +10 cm is:
\[ \Delta t = t_2 - t_1 = 7 - 1 = 6 \, \text{s} \]

### Step 9: Minimum time required
Since we need the minimum time to move between these two points, we take half of this time (as the particle moves to the first point and then back to the second):
\[ \text{Minimum time} = \frac{6}{2} = 3 \, \text{s} \]

### Final Answer
The minimum time required for the particle to move between two points 10 cm on either side of the mean position is **2 seconds**.

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