A particle executes a linear simple harmonic motion of amplitude 20 cm and time period 4 s. The minimum time period required for the particle to move between two points located at 10 cm on either side of equilibrium position is t seconds, then compute 3t + 2 .

Explanation

To solve the problem step by step, we need to find the minimum time \( t \) required for a particle executing simple harmonic motion (SHM) to move between two points located at 10 cm on either side of the equilibrium position. The amplitude is given as 20 cm, and the time period is 4 seconds.

### Step 1: Understand the Motion
The particle oscillates between -20 cm and +20 cm, with the equilibrium position at 0 cm. The points we are interested in are -10 cm and +10 cm.

### Step 2: Determine Angular Frequency
The angular frequency \( \omega \) can be calculated using the formula:
\[ \omega = \frac{2\pi}{T} \]
where \( T \) is the time period. Given \( T = 4 \) seconds:
\[ \omega = \frac{2\pi}{4} = \frac{\pi}{2} \, \text{rad/s} \]

### Step 3: Write the Equation of Motion
The displacement \( x \) of the particle in SHM can be expressed as:
\[ x = A \sin(\omega t + \phi) \]
where \( A \) is the amplitude and \( \phi \) is the phase constant. Here, \( A = 20 \) cm. Assuming the motion starts at the maximum displacement, we can set \( \phi = \frac{\pi}{2} \):
\[ x = 20 \sin\left(\frac{\pi}{2} t + \frac{\pi}{2}\right) \]
This simplifies to:
\[ x = 20 \cos\left(\frac{\pi}{2} t\right) \]

### Step 4: Find Time to Reach +10 cm
To find the time \( t_1 \) when the particle reaches +10 cm:
\[ 10 = 20 \cos\left(\frac{\pi}{2} t_1\right) \]
\[ \cos\left(\frac{\pi}{2} t_1\right) = \frac{1}{2} \]
The angle whose cosine is \( \frac{1}{2} \) is \( \frac{\pi}{3} \):
\[ \frac{\pi}{2} t_1 = \frac{\pi}{3} \implies t_1 = \frac{2}{3} \, \text{s} \]

### Step 5: Find Time to Reach -10 cm
Next, we find the time \( t_2 \) when the particle reaches -10 cm:
\[ -10 = 20 \cos\left(\frac{\pi}{2} t_2\right) \]
\[ \cos\left(\frac{\pi}{2} t_2\right) = -\frac{1}{2} \]
The angle whose cosine is \( -\frac{1}{2} \) is \( \frac{2\pi}{3} \):
\[ \frac{\pi}{2} t_2 = \frac{2\pi}{3} \implies t_2 = \frac{4}{3} \, \text{s} \]

### Step 6: Calculate Minimum Time \( t \)
The minimum time \( t \) required to move from +10 cm to -10 cm is:
\[ t = t_2 - t_1 = \frac{4}{3} - \frac{2}{3} = \frac{2}{3} \, \text{s} \]

### Step 7: Compute \( 3t + 2 \)
Now, we compute \( 3t + 2 \):
\[ 3t + 2 = 3 \left(\frac{2}{3}\right) + 2 = 2 + 2 = 4 \]

### Final Answer
Thus, the final answer is:
\[ \boxed{4} \]