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 follow these instructions:

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

### Step 2: Identify the positions of interest
The particle moves between two points that are 10 cm on either side of the mean position. Therefore, the positions are:
- Point A = -10 cm (10 cm left of the mean position)
- Point B = +10 cm (10 cm right of the mean position)

### Step 3: Write the equation of motion
The equation for simple harmonic motion (SHM) is given by:
\[ x(t) = A \sin(\omega t) \]
Where:
- \( \omega \) (angular frequency) is related to the time period by:
\[ \omega = \frac{2\pi}{T} \]

### Step 4: Calculate the angular frequency
Using the time period:
\[ \omega = \frac{2\pi}{12} = \frac{\pi}{6} \, \text{rad/s} \]

### Step 5: Set up the equation for the position
Now, we need to find the time \( t \) when the particle is at position \( x = 10 \) cm:
\[ 10 = 20 \sin\left(\frac{\pi}{6} t\right) \]

### Step 6: Solve for \( \sin\left(\frac{\pi}{6} t\right) \)
Dividing both sides by 20:
\[ \sin\left(\frac{\pi}{6} t\right) = \frac{10}{20} = \frac{1}{2} \]

### Step 7: Find the angle corresponding to \( \sin\left(\frac{\pi}{6} t\right) = \frac{1}{2} \)
The sine function equals \( \frac{1}{2} \) at:
\[ \frac{\pi}{6} t = \frac{\pi}{6} \quad \text{(for } t = 1 \text{ s)} \]
and also at:
\[ \frac{\pi}{6} t = \frac{5\pi}{6} \quad \text{(for } t = 5 \text{ s)} \]

### Step 8: Determine the time to move from A to B
The particle takes 1 second to move from the mean position to point A (10 cm) and another 1 second to move from point A to point B (10 cm). Therefore, the total time to move from A to B is:
\[ \text{Total time} = 1 \, \text{s} + 1 \, \text{s} = 2 \, \text{s} \]

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

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