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A closed organ pipe and an open organ pi...

A closed organ pipe and an open organ pipe of same length produce 2 beats/second while vibrating in their fundamental modes. The length of the open organ pipe is halved and that of closed pipe is doubled. Then, the number of beats produced per second while vibrating in the fundamental mode is

A

2

B

6

C

8

D

7

Text Solution

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The correct Answer is:
To solve the problem, we need to determine the number of beats produced per second when the lengths of the open and closed organ pipes are modified. Let's break it down step by step. ### Step 1: Understand the initial conditions We know that: - A closed organ pipe and an open organ pipe of the same length produce 2 beats/second while vibrating in their fundamental modes. Let: - \( f_C \) = fundamental frequency of the closed organ pipe - \( f_B \) = fundamental frequency of the open organ pipe The number of beats per second is given by: \[ f_B - f_C = 2 \text{ Hz} \] ### Step 2: Write the formulas for the fundamental frequencies The fundamental frequency for the closed organ pipe is given by: \[ f_C = \frac{V}{4L} \] The fundamental frequency for the open organ pipe is given by: \[ f_B = \frac{V}{2L} \] ### Step 3: Set up the equation for beats From the information given: \[ \frac{V}{2L} - \frac{V}{4L} = 2 \] Simplifying this: \[ \frac{2V}{4L} - \frac{V}{4L} = 2 \] \[ \frac{V}{4L} = 2 \] Thus, we can express \( \frac{V}{L} \): \[ \frac{V}{L} = 8 \text{ Hz} \] ### Step 4: Modify the lengths of the pipes Now, we need to consider the new lengths: - The length of the open organ pipe is halved: \( L_{open} = \frac{L}{2} \) - The length of the closed organ pipe is doubled: \( L_{closed} = 2L \) ### Step 5: Calculate the new frequencies For the new open pipe: \[ f_B' = \frac{V}{2 \cdot \frac{L}{2}} = \frac{V}{L} \] For the new closed pipe: \[ f_C' = \frac{V}{4 \cdot 2L} = \frac{V}{8L} \] ### Step 6: Calculate the new beat frequency Now we can find the new beat frequency: \[ f_B' - f_C' = \frac{V}{L} - \frac{V}{8L} \] Simplifying this: \[ = \frac{8V}{8L} - \frac{V}{8L} = \frac{7V}{8L} \] ### Step 7: Substitute the value of \( \frac{V}{L} \) We know \( \frac{V}{L} = 8 \text{ Hz} \): \[ f_B' - f_C' = \frac{7 \cdot 8}{8} = 7 \text{ Hz} \] ### Conclusion The number of beats produced per second while vibrating in the fundamental mode after the changes is: \[ \text{Number of beats} = 7 \text{ Hz} \] ### Final Answer Thus, the correct option is **D) 7**. ---

To solve the problem, we need to determine the number of beats produced per second when the lengths of the open and closed organ pipes are modified. Let's break it down step by step. ### Step 1: Understand the initial conditions We know that: - A closed organ pipe and an open organ pipe of the same length produce 2 beats/second while vibrating in their fundamental modes. Let: - \( f_C \) = fundamental frequency of the closed organ pipe ...
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