In a medium in which a transverse prograssive wave is travelling, the phase difference between two points with a separation of 1.25 cm is `(pi//4). If the frequancy of wave is 1000 Hz. Its velocity will be

Options

A. `10^(4) m//s`
B. `125 m//s`
C. `100 m//s`
D. `10 m//s`

Explanation

To solve the problem, we need to find the velocity of a transverse progressive wave given the phase difference, separation between two points, and frequency. Here's the step-by-step solution:

### Step 1: Understand the given data
- Phase difference (ΔΦ) = π/4
- Separation between two points (Δx) = 1.25 cm = 1.25 × 10^(-2) m
- Frequency (f) = 1000 Hz

### Step 2: Use the relationship between phase difference and path difference
The relationship between phase difference (ΔΦ) and path difference (Δx) is given by:
\[ \Delta \Phi = k \cdot \Delta x \]
where \( k \) is the wave number.

### Step 3: Express wave number in terms of wavelength
The wave number \( k \) is related to the wavelength (λ) by:
\[ k = \frac{2\pi}{\lambda} \]
Substituting this into the phase difference equation gives:
\[ \Delta \Phi = \frac{2\pi}{\lambda} \cdot \Delta x \]

### Step 4: Rearrange to find the wavelength
Rearranging the equation to solve for λ:
\[ \lambda = \frac{2\pi \cdot \Delta x}{\Delta \Phi} \]

### Step 5: Substitute the known values
Substituting the known values into the equation:
\[ \lambda = \frac{2\pi \cdot (1.25 \times 10^{-2})}{\frac{\pi}{4}} \]

### Step 6: Simplify the expression
The π in the numerator and denominator cancels out:
\[ \lambda = \frac{2 \cdot (1.25 \times 10^{-2}) \cdot 4}{1} = 10 \times 10^{-2} = 0.1 \text{ m} \]

### Step 7: Calculate the velocity of the wave
The velocity (v) of the wave is given by the formula:
\[ v = f \cdot \lambda \]
Substituting the values:
\[ v = 1000 \cdot 0.1 = 100 \text{ m/s} \]

### Final Answer
The velocity of the wave is **100 m/s**.

---