The displacement of a progressive wave is represented by `y=Asin(omegat-kx)` where x is distance and t is time. The dimensions of `(omega)/(k)` are same as those of

Options

A. velocity
B. wave number
C. wavelength
D. frequency

Explanation

To solve the question, we need to analyze the given equation for the displacement of a progressive wave, which is represented as:

\[ y = A \sin(\omega t - kx) \]

where:
- \( y \) is the displacement,
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( k \) is the wave number,
- \( x \) is the distance,
- \( t \) is the time.

We are asked to find the dimensions of the ratio \( \frac{\omega}{k} \) and determine what physical quantity it corresponds to.

### Step 1: Understand the terms involved
The term \( \omega t - kx \) must be dimensionless because it represents an angle in the sine function. This means that the dimensions of \( \omega t \) must equal the dimensions of \( kx \).

### Step 2: Find the dimensions of \( \omega \)
The angular frequency \( \omega \) is defined as:

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

where \( T \) is the time period. Thus, the dimensions of \( \omega \) can be expressed as:

\[ [\omega] = [T]^{-1} = T^{-1} \]

### Step 3: Find the dimensions of \( k \)
The wave number \( k \) is defined as:

\[ k = \frac{2\pi}{\lambda} \]

where \( \lambda \) is the wavelength. The dimensions of \( k \) can be expressed as:

\[ [k] = [\lambda]^{-1} = L^{-1} \]

### Step 4: Calculate the dimensions of \( \frac{\omega}{k} \)
Now we can find the dimensions of the ratio \( \frac{\omega}{k} \):

\[ \frac{\omega}{k} = \frac{T^{-1}}{L^{-1}} = T^{-1} \cdot L = \frac{L}{T} \]

### Step 5: Identify the physical quantity
The dimensions \( \frac{L}{T} \) correspond to the dimensions of velocity. Therefore, we conclude that:

\[ \frac{\omega}{k} \text{ has the same dimensions as velocity.} \]

### Final Answer
The dimensions of \( \frac{\omega}{k} \) are the same as those of velocity.

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