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An open pipe is suddenly closed at one e...

An open pipe is suddenly closed at one end with the result that the frequency of third harmonic of the closed pipe is found to be higher by `100 Hz` then the fundamental frequency of the open pipe. The fundamental frequency of the open pipe is

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To solve the problem, we need to find the fundamental frequency of the open pipe given the information about the closed pipe. ### Step-by-Step Solution: 1. **Understanding the Frequencies**: - Let \( f \) be the fundamental frequency of the open pipe. - The frequency of the third harmonic of the closed pipe is given to be \( f' \). - We know that \( f' = f + 100 \, \text{Hz} \). 2. **Fundamental Frequency of the Open Pipe**: - For an open pipe, the fundamental frequency is given by: \[ f = \frac{v}{\lambda} \] where \( \lambda \) is the wavelength and \( v \) is the speed of sound in air. - The wavelength for the fundamental frequency of an open pipe is \( \lambda = 2L \), where \( L \) is the length of the pipe. - Thus, we have: \[ f = \frac{v}{2L} \] 3. **Third Harmonic of the Closed Pipe**: - For a closed pipe, the third harmonic corresponds to the frequency where there are 3/2 wavelengths fitting into the length of the pipe. - The wavelength for the third harmonic of a closed pipe is given by: \[ \lambda' = \frac{4L}{3} \] - Therefore, the frequency of the third harmonic is: \[ f' = \frac{v}{\lambda'} = \frac{v}{\frac{4L}{3}} = \frac{3v}{4L} \] 4. **Setting Up the Equation**: - From the information given, we have: \[ f' = f + 100 \] - Substituting the expressions for \( f \) and \( f' \): \[ \frac{3v}{4L} = \frac{v}{2L} + 100 \] 5. **Solving for \( v \)**: - To eliminate \( v \), we can multiply the entire equation by \( 4L \): \[ 3v = 2v + 400L \] - Rearranging gives: \[ 3v - 2v = 400L \implies v = 400L \] 6. **Finding the Fundamental Frequency**: - Now substituting \( v = 400L \) back into the formula for \( f \): \[ f = \frac{v}{2L} = \frac{400L}{2L} = 200 \, \text{Hz} \] ### Final Answer: The fundamental frequency of the open pipe is \( \boxed{200 \, \text{Hz}} \).

To solve the problem, we need to find the fundamental frequency of the open pipe given the information about the closed pipe. ### Step-by-Step Solution: 1. **Understanding the Frequencies**: - Let \( f \) be the fundamental frequency of the open pipe. - The frequency of the third harmonic of the closed pipe is given to be \( f' \). - We know that \( f' = f + 100 \, \text{Hz} \). ...
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