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In an organ pipe the distance between th...

In an organ pipe the distance between the adjacent node is `4 cm`. Find the frequency of source if speed of sound in air is `336 m//s`.

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To solve the problem step by step, we need to find the frequency of the sound produced by an organ pipe given the distance between adjacent nodes and the speed of sound in air. ### Step-by-Step Solution: 1. **Identify the distance between adjacent nodes**: The problem states that the distance between adjacent nodes is `4 cm`. 2. **Convert the distance to meters**: Since the speed of sound is given in meters per second, we need to convert `4 cm` to meters: \[ 4 \, \text{cm} = 4 \times 10^{-2} \, \text{m} = 0.04 \, \text{m} \] 3. **Relate the distance between nodes to the wavelength**: In a standing wave, the distance between two adjacent nodes is equal to half the wavelength (\(\lambda\)): \[ \text{Distance between nodes} = \frac{\lambda}{2} \] Therefore, we can set up the equation: \[ 0.04 \, \text{m} = \frac{\lambda}{2} \] 4. **Solve for the wavelength (\(\lambda\))**: To find \(\lambda\), we can rearrange the equation: \[ \lambda = 2 \times 0.04 \, \text{m} = 0.08 \, \text{m} = 8 \times 10^{-2} \, \text{m} \] 5. **Use the speed of sound to find frequency**: The frequency (\(f\)) can be calculated using the formula: \[ f = \frac{v}{\lambda} \] where \(v\) is the speed of sound. Given that \(v = 336 \, \text{m/s}\): \[ f = \frac{336 \, \text{m/s}}{0.08 \, \text{m}} \] 6. **Calculate the frequency**: Performing the division: \[ f = \frac{336}{0.08} = 4200 \, \text{Hz} \] 7. **Convert frequency to kilohertz**: To convert Hertz to kilohertz, divide by 1000: \[ f = \frac{4200 \, \text{Hz}}{1000} = 4.2 \, \text{kHz} \] ### Final Answer: The frequency of the source is \(4.2 \, \text{kHz}\). ---

To solve the problem step by step, we need to find the frequency of the sound produced by an organ pipe given the distance between adjacent nodes and the speed of sound in air. ### Step-by-Step Solution: 1. **Identify the distance between adjacent nodes**: The problem states that the distance between adjacent nodes is `4 cm`. 2. **Convert the distance to meters**: ...
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RESONANCE ENGLISH-SOUND WAVES-Exercise- 1 PART - I
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  3. The equation of a longitudinal standing wave due to superposition of t...

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  4. A closed organ pipe has length L. The air in it is vibrating in third ...

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  5. The speed of sound in an air column of 80 cm closed at one end is 320 ...

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  6. In an organ pipe the distance between the adjacent node is 4 cm. Find ...

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  7. Two pipes P(1) and P(2) are closed and open respectively. P(1) has a l...

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  8. Two successive resonance frequencies in an open organ pipe are 1944 Hz...

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  9. A closed organ pipe of length l = 100 cm is cut into two unequal piece...

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  10. Find the number of possible natural oscillations of air column in a pi...

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  15. An observer rides with a sound source of frequency f and moving with v...

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  16. A stationary source sends forth monochromatic sound. A wall approaches...

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  17. A source of sonic oscillations with frequency n(0)=600Hz moves away an...

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  18. A sound wave of frequency f propagating through air with a velocity c,...

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  19. two trains move towards each other sith the same speed. The speed of s...

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  20. A tuning fork P of unknows frequency gives 7 beats in 2 seconds with a...

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