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A tuning fork A of frequency 512 Hz soun...

A tuning fork A of frequency 512 Hz sounded with another tuning fork B gives 5 beats/sec. On loading B with a piece of wax, and on sounding, again 5 beats/sec are produced. Calculate the frequency of fork B.

A

517 Hz

B

500 Hz

C

600 Hz

D

567 Hz

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The correct Answer is:
To solve the problem, we need to determine the frequency of tuning fork B (fB) given that tuning fork A has a frequency of 512 Hz and they produce 5 beats per second. ### Step-by-Step Solution: 1. **Understanding Beat Frequency**: The beat frequency is the absolute difference between the frequencies of two tuning forks. The formula for beat frequency (f_beat) is: \[ f_{beat} = |f_A - f_B| \] Given that \( f_A = 512 \) Hz and \( f_{beat} = 5 \) Hz, we can express this as: \[ |512 - f_B| = 5 \] 2. **Setting Up the Equations**: This absolute value equation gives us two possible scenarios: - Case 1: \( 512 - f_B = 5 \) - Case 2: \( f_B - 512 = 5 \) From Case 1: \[ f_B = 512 - 5 = 507 \text{ Hz} \] From Case 2: \[ f_B = 512 + 5 = 517 \text{ Hz} \] So, the two potential frequencies for tuning fork B are 507 Hz and 517 Hz. 3. **Analyzing the Effect of Wax**: When tuning fork B is loaded with wax, its frequency will decrease. If we assume that the original frequency of B was 517 Hz, then after loading with wax, it will decrease further. If \( f_B = 517 \) Hz, after loading with wax, the frequency will drop below 517 Hz, and we can check if it can still produce 5 beats per second with tuning fork A: \[ |512 - f_B'| = 5 \] where \( f_B' < 517 \). 4. **Verifying the Frequencies**: - If \( f_B = 507 \) Hz, after loading with wax, \( f_B \) would decrease further, making it impossible to maintain a beat frequency of 5 Hz with 512 Hz. - If \( f_B = 517 \) Hz, after loading with wax, \( f_B \) will decrease, and it can still maintain a beat frequency of 5 Hz as it approaches 512 Hz. 5. **Conclusion**: Therefore, the only viable frequency for tuning fork B that satisfies both conditions (before and after loading with wax) is: \[ f_B = 517 \text{ Hz} \] ### Final Answer: The frequency of tuning fork B is **517 Hz**. ---

To solve the problem, we need to determine the frequency of tuning fork B (fB) given that tuning fork A has a frequency of 512 Hz and they produce 5 beats per second. ### Step-by-Step Solution: 1. **Understanding Beat Frequency**: The beat frequency is the absolute difference between the frequencies of two tuning forks. The formula for beat frequency (f_beat) is: \[ f_{beat} = |f_A - f_B| ...
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