A particle is performing SHM according to the equation `x=(3cm)sin((2pit)/(18)+(pi)/(6))`, where t is in seconds. The distance travelled by the particle in 39 s is

Options

A. `24 cm`
B. `1.5 cm`
C. `25.5 cm`
D. none of these

Explanation

To solve the problem of finding the distance traveled by a particle performing simple harmonic motion (SHM) given the equation:

\[ x = (3 \, \text{cm}) \sin\left(\frac{2\pi t}{18} + \frac{\pi}{6}\right) \]

we will follow these steps:

### Step 1: Identify the parameters of the SHM
From the equation, we can identify the amplitude \( A \) and the angular frequency \( \omega \).

- Amplitude \( A = 3 \, \text{cm} \)
- Angular frequency \( \omega = \frac{2\pi}{18} \, \text{rad/s} \)

### Step 2: Calculate the time period \( T \)
The time period \( T \) of the motion can be calculated using the formula:

\[ T = \frac{2\pi}{\omega} \]

Substituting the value of \( \omega \):

\[ T = \frac{2\pi}{\frac{2\pi}{18}} = 18 \, \text{s} \]

### Step 3: Determine the number of complete cycles in 39 seconds
To find out how many complete cycles the particle completes in 39 seconds, we divide the total time by the time period:

\[ \text{Number of cycles} = \frac{39 \, \text{s}}{T} = \frac{39}{18} = 2.1667 \]

This means the particle completes 2 full cycles and a fraction of the third cycle.

### Step 4: Calculate the distance traveled in complete cycles
In one complete cycle, the particle travels a distance of \( 4A \) (from the mean position to the maximum amplitude, back to the mean position, to the minimum amplitude, and back to the mean position):

\[ \text{Distance in one cycle} = 4A = 4 \times 3 \, \text{cm} = 12 \, \text{cm} \]

Thus, for 2 complete cycles:

\[ \text{Distance for 2 cycles} = 2 \times 12 \, \text{cm} = 24 \, \text{cm} \]

### Step 5: Calculate the distance traveled in the remaining time
Now, we need to calculate the distance traveled in the remaining \( 0.1667 \) of a cycle.

The fraction of the cycle corresponds to:

\[ 0.1667 \times T = 0.1667 \times 18 \approx 3 \, \text{s} \]

### Step 6: Find the position at \( t = 3 \, \text{s} \)
Using the original equation, we can find the position at \( t = 3 \, \text{s} \):

\[ x = 3 \sin\left(\frac{2\pi \times 3}{18} + \frac{\pi}{6}\right) = 3 \sin\left(\frac{\pi}{3} + \frac{\pi}{6}\right) = 3 \sin\left(\frac{2\pi}{6} + \frac{\pi}{6}\right) = 3 \sin\left(\frac{3\pi}{6}\right) = 3 \sin\left(\frac{\pi}{2}\right) = 3 \, \text{cm} \]

### Step 7: Calculate the distance for the remaining part of the cycle
In the remaining \( 3 \, \text{s} \), the particle moves from the maximum position \( 3 \, \text{cm} \) back to the mean position \( 0 \, \text{cm} \) and then to the minimum position \( -3 \, \text{cm} \). The total distance traveled in this part is:

\[ 3 \, \text{cm} (to the mean) + 3 \, \text{cm} (to the minimum) = 6 \, \text{cm} \]

### Step 8: Total distance traveled
Finally, we sum the distances:

\[ \text{Total distance} = 24 \, \text{cm} + 6 \, \text{cm} = 30 \, \text{cm} \]

### Final Answer
The distance traveled by the particle in 39 seconds is:

\[ \boxed{30 \, \text{cm}} \]