A progressive and a stationary simple harmonic wave each has the same frequency 250Hz and the same velocity of 30m/s. Calculate the distance between consecutive nodes (in cm) in the stationary wave.

Answer

To solve the problem, we need to find the distance between consecutive nodes in a stationary wave given the frequency and velocity of the wave.

### Step-by-Step Solution:

1. **Identify the given values:**
- Frequency (f) = 250 Hz
- Velocity (v) = 30 m/s

2. **Use the wave equation to find the wavelength (λ):**
The relationship between wave speed (v), frequency (f), and wavelength (λ) is given by the formula:
\[ v = f \cdot \lambda \]
Rearranging this formula to solve for wavelength gives us:
\[ \lambda = \frac{v}{f} \]

3. **Substitute the values into the formula:**
\[ \lambda = \frac{30 \, \text{m/s}}{250 \, \text{Hz}} = \frac{30}{250} = \frac{3}{25} \, \text{m} \]

4. **Calculate the distance between consecutive nodes:**
In a stationary wave, the distance between two consecutive nodes is half the wavelength:
\[ \text{Distance between nodes} = \frac{\lambda}{2} \]
Substituting the value of λ we found:
\[ \text{Distance between nodes} = \frac{3/25}{2} = \frac{3}{50} \, \text{m} \]

5. **Convert the distance from meters to centimeters:**
Since 1 meter = 100 centimeters, we convert the distance:
\[ \frac{3}{50} \, \text{m} = \frac{3}{50} \times 100 \, \text{cm} = 6 \, \text{cm} \]

### Final Answer:
The distance between consecutive nodes in the stationary wave is **6 cm**.