When a harmonic wave is proppagating through a medium, the displacement 'y' of a particle of the medium is represented by `y = `10 sin (2pi)/(5)` (1800t- x). The time period will be

Options

A. 1/360 sec
B. 1/36 sec
C. 36 sec
D. 360 sec

Answer

To find the time period of the harmonic wave represented by the equation \( y = 10 \sin\left(\frac{2\pi}{5}(1800t - x)\right) \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the General Form of the Wave Equation**:
The general form of a harmonic wave is given by:
\[ y = A \sin(\omega t - kx) \]
where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( k \) is the wave number.

2. **Compare with the Given Equation**:
The given equation is:
\[ y = 10 \sin\left(\frac{2\pi}{5}(1800t - x)\right) \]
We can rewrite this as:
\[ y = 10 \sin\left( \left(\frac{2\pi}{5} \cdot 1800\right)t - \frac{2\pi}{5}x\right) \]
From this, we can identify:
- \( \omega = \frac{2\pi}{5} \cdot 1800 \)

3. **Calculate the Angular Frequency (\(\omega\))**:
Calculate \( \omega \):
\[ \omega = \frac{2\pi}{5} \cdot 1800 = \frac{3600\pi}{5} = 720\pi \text{ rad/s} \]

4. **Relate Angular Frequency to Frequency**:
The relationship between angular frequency (\(\omega\)) and frequency (\(f\)) is given by:
\[ \omega = 2\pi f \]
Therefore, we can find \( f \):
\[ f = \frac{\omega}{2\pi} = \frac{720\pi}{2\pi} = 360 \text{ Hz} \]

5. **Calculate the Time Period (T)**:
The time period \( T \) is the reciprocal of the frequency:
\[ T = \frac{1}{f} = \frac{1}{360} \text{ seconds} \]

### Final Answer:
The time period \( T \) of the wave is:
\[ T = \frac{1}{360} \text{ seconds} \]