The equation for stationary wave is `y = 0.005 cos (62.8t + 3.14x + (pi)/(3))` its periodic time T and wavelenggth`lambda` are -

Options

A. 3.14 sec, 1m
B. 1 sec , 1m
C. 10 sec, 2m
D. 0.1 sec, 1m

Answer

To find the periodic time \( T \) and the wavelength \( \lambda \) from the given equation of a stationary wave:

### Step 1: Identify the parameters from the wave equation
The given wave equation is:
\[ y = 0.005 \cos(62.8t + 3.14x + \frac{\pi}{3}) \]
From this equation, we can identify:
- Angular frequency \( \omega = 62.8 \) rad/s
- Wave number \( k = 3.14 \) rad/m

### Step 2: Calculate the periodic time \( T \)
The relationship between angular frequency \( \omega \) and the time period \( T \) is given by:
\[ \omega = \frac{2\pi}{T} \]
Rearranging this gives:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{62.8} \]
Using \( \pi \approx 3.14 \):
\[ T = \frac{2 \times 3.14}{62.8} = \frac{6.28}{62.8} \approx 0.1 \text{ seconds} \]

### Step 3: Calculate the wavelength \( \lambda \)
The relationship between wave number \( k \) and wavelength \( \lambda \) is given by:
\[ k = \frac{2\pi}{\lambda} \]
Rearranging this gives:
\[ \lambda = \frac{2\pi}{k} \]
Substituting the value of \( k \):
\[ \lambda = \frac{2\pi}{3.14} \]
Using \( \pi \approx 3.14 \):
\[ \lambda = \frac{2 \times 3.14}{3.14} = 2 \text{ meters} \]

### Final Results
- Periodic time \( T \approx 0.1 \) seconds
- Wavelength \( \lambda = 2 \) meters

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