A transformer is humming at a frequency of 6000 Hz and produces a standing wave in air. What is the distance between successive nodes? What is the distance between successive antinodes? If the speed of sound close to the standard value of 340 m/s.

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### Understanding Standing Waves Produced by a Transformer

A transformer is humming at a frequency of 600 Hz and produces a standing wave in air. This raises an interesting question relevant to wave physics:

**What is the distance between successive nodes? What is the distance between successive antinodes?**

Assuming the speed of sound is close to the standard value of 340 m/s, we can explore how these distances can be calculated.

#### Key Concepts:

1. **Standing Waves**:
   - **Node (N)**: A point in the medium where the displacement is always zero.
   - **Antinode (A)**: A point where the displacement reaches its maximum value.

2. **Wavelength (λ)**:
   - The distance between two consecutive nodes or two consecutive antinodes. 

3. **Speed of Sound (v)**:
   - Given as 340 m/s in this problem.

#### Calculation Steps:

1. **Determine Wavelength**:
   Using the formula for the speed of sound (v) in relation to frequency (f) and wavelength (λ):
   \[ v = f \times \lambda \]
   Solving for wavelength (λ):
   \[ \lambda = \frac{v}{f} \]

   Given:
   - \( v = 340 \, \text{m/s} \)
   - \( f = 600 \, \text{Hz} \)

   So:
   \[ \lambda = \frac{340 \, \text{m/s}}{600 \, \text{Hz}} \]
   \[ \lambda = \frac{340}{600} \, \text{m} \]
   \[ \lambda = 0.5667 \, \text{m} \]

2. **Distance between successive nodes**:
   The distance between two successive nodes is half of the wavelength:
   \[ \text{Distance between successive nodes} = \frac{\lambda}{2} \]
   \[ = \frac{0.5667 \, \text{m}}{2} \]
   \[ = 0.28335 \, \text{m} \]

3. **Distance between successive antinodes**:
   Similarly, the distance between successive antinodes is also half of the wavelength:
   \[ \text{Distance between successive antinodes} = \frac{\lambda}{2} \]
Transcribed Image Text:### Understanding Standing Waves Produced by a Transformer A transformer is humming at a frequency of 600 Hz and produces a standing wave in air. This raises an interesting question relevant to wave physics: **What is the distance between successive nodes? What is the distance between successive antinodes?** Assuming the speed of sound is close to the standard value of 340 m/s, we can explore how these distances can be calculated. #### Key Concepts: 1. **Standing Waves**: - **Node (N)**: A point in the medium where the displacement is always zero. - **Antinode (A)**: A point where the displacement reaches its maximum value. 2. **Wavelength (λ)**: - The distance between two consecutive nodes or two consecutive antinodes. 3. **Speed of Sound (v)**: - Given as 340 m/s in this problem. #### Calculation Steps: 1. **Determine Wavelength**: Using the formula for the speed of sound (v) in relation to frequency (f) and wavelength (λ): \[ v = f \times \lambda \] Solving for wavelength (λ): \[ \lambda = \frac{v}{f} \] Given: - \( v = 340 \, \text{m/s} \) - \( f = 600 \, \text{Hz} \) So: \[ \lambda = \frac{340 \, \text{m/s}}{600 \, \text{Hz}} \] \[ \lambda = \frac{340}{600} \, \text{m} \] \[ \lambda = 0.5667 \, \text{m} \] 2. **Distance between successive nodes**: The distance between two successive nodes is half of the wavelength: \[ \text{Distance between successive nodes} = \frac{\lambda}{2} \] \[ = \frac{0.5667 \, \text{m}}{2} \] \[ = 0.28335 \, \text{m} \] 3. **Distance between successive antinodes**: Similarly, the distance between successive antinodes is also half of the wavelength: \[ \text{Distance between successive antinodes} = \frac{\lambda}{2} \]
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