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two pipes have each of length 2 m, one i...

two pipes have each of length 2 m, one is closed at on end and the other is open at both ends. The speed of sound in air is 340 m/s . The frequency at which both can resonate is ?

A

340 Hz

B

510 Hz

C

42.5 Hz

D

none of these

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of finding the resonant frequencies of two pipes (one closed at one end and the other open at both ends), we can follow these steps: ### Step 1: Understand the properties of the pipes - **Pipe 1** (closed at one end): The fundamental frequency (first harmonic) is given by the formula: \[ f_1 = \frac{v}{4L} \] where \( v \) is the speed of sound and \( L \) is the length of the pipe. The harmonics for this pipe will be odd multiples of the fundamental frequency: \( f_1, 3f_1, 5f_1, \ldots \). - **Pipe 2** (open at both ends): The fundamental frequency is given by: \[ f_2 = \frac{v}{2L} \] The harmonics for this pipe will include all integer multiples of the fundamental frequency: \( f_2, 2f_2, 3f_2, \ldots \). ### Step 2: Calculate the fundamental frequencies Given: - Length of each pipe, \( L = 2 \, \text{m} \) - Speed of sound, \( v = 340 \, \text{m/s} \) For **Pipe 1**: \[ f_1 = \frac{340}{4 \times 2} = \frac{340}{8} = 42.5 \, \text{Hz} \] For **Pipe 2**: \[ f_2 = \frac{340}{2 \times 2} = \frac{340}{4} = 85 \, \text{Hz} \] ### Step 3: Determine the resonant frequencies The resonant frequencies for Pipe 1 (closed at one end) are: \[ f_{1n} = (2n - 1)f_1 = (2n - 1) \times 42.5 \, \text{Hz} \quad (n = 1, 2, 3, \ldots) \] The resonant frequencies for Pipe 2 (open at both ends) are: \[ f_{2m} = mf_2 = m \times 85 \, \text{Hz} \quad (m = 1, 2, 3, \ldots) \] ### Step 4: Set the equations equal to find common frequencies To find the resonant frequencies where both pipes resonate, we set: \[ (2n - 1) \times 42.5 = m \times 85 \] ### Step 5: Solve for integers \( n \) and \( m \) Rearranging gives: \[ 2n - 1 = \frac{m \times 85}{42.5} = 2m \] This implies: \[ 2n - 1 = 2m \implies 2n = 2m + 1 \implies n = m + \frac{1}{2} \] Since \( n \) and \( m \) must be natural numbers, this equation cannot hold true. Therefore, there are no integer solutions for \( n \) and \( m \). ### Conclusion The two pipes do not resonate at any common frequency. ### Final Answer The answer is: **They do not resonate at any frequency.** ---
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