The equation of a plane progressive wave travelling along positive direction of `x-`axis is `y=r sin [(2pit)/(T)-(2pix)/(lambda)]` where `y=` displacement of particle at `(x,t),r=` amplitude of vibratio of particle, `T=` time period of wave motion, `lambda=` wavelength of wave ,` x=` starting distance of wave from the origin. Velocity of wave,
`upsilon=vlambda=(lambda)/(T)=` constant.
Acceleration of wave, `a=0`.
Velocity of particle at time `t=(dy)/(dt)`
Acceleration of particle at time `t=(d^(2)y)/(dt^(2))`
Time period of wave motion is

Options

A. `pi//250s`
B. `pi//500s`
C. `250//pi s`
D. `500//pis`

Explanation

To solve the problem, we need to find the time period \( T \) of the wave motion from the given wave equation:

\[ y = r \sin \left( \frac{2\pi t}{T} - \frac{2\pi x}{\lambda} \right) \]

### Step 1: Identify the coefficients in the wave equation
The given wave equation can be compared with the standard form of a progressive wave. The coefficient of \( t \) in the sine function is \( \frac{2\pi}{T} \) and the coefficient of \( x \) is \( -\frac{2\pi}{\lambda} \).

### Step 2: Extract the coefficient of \( t \)
From the video transcript, we see that the equation is given as:

\[ y = 0.25 \times 10^{-3} \sin(500t - 0.025x) \]

Here, we can identify that the coefficient of \( t \) is \( 500 \).

### Step 3: Set up the equation for \( T \)
From the comparison, we have:

\[ \frac{2\pi}{T} = 500 \]

### Step 4: Solve for \( T \)
To find \( T \), we rearrange the equation:

\[ T = \frac{2\pi}{500} \]

### Step 5: Calculate \( T \)
Now, we can simplify this:

\[ T = \frac{2\pi}{500} = \frac{\pi}{250} \]

### Conclusion
Thus, the time period of the wave motion is:

\[ T = \frac{\pi}{250} \text{ seconds} \]