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When a tuning fork of frequency 341 Hz is sounded with another tuning fork, six beats per second are heard. When the second tuning fork is loaded with wax and sounded with the first tuning fork, the number of beats is two per second. The natural frequency of the second tuning fork is
  • a)
    334
  • b)
    339
  • c)
    343
  • d)
    347
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
When a tuning fork of frequency 341 Hzis sounded with another tuning f...
Analysis:
The beat frequency heard when two tuning forks are sounded together is given by the difference in their frequencies. Let's denote the natural frequency of the first tuning fork as \( f_1 = 341 \, \text{Hz} \).

Given Information:
- Beat frequency when two forks are sounded together: 6 beats/s
- Beat frequency when the loaded fork is sounded with the first fork: 2 beats/s

Solution:

Calculating the frequency of the second tuning fork:
- When two forks are sounded together, the beat frequency is 6 Hz. This implies that the difference in their frequencies is 6 Hz.
\( |f_1 - f_2| = 6 \)
- When the loaded fork is sounded with the first fork, the beat frequency is 2 Hz. This implies that the difference in their frequencies is 2 Hz.
\( |f_1 - f_2'| = 2 \)
- Subtracting the two equations, we get:
\( f_2 - f_2' = 4 \)
- Adding the above two equations, we get:
\( f_2 + f_2' = 8 \)
- Solving the above two equations, we get:
\( f_2 = 6 \, \text{Hz} \)
\( f_2' = 2 \, \text{Hz} \)

Conclusion:
The natural frequency of the second tuning fork is \( 347 \, \text{Hz} \), which is option D.
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When a tuning fork of frequency 341 Hzis sounded with another tuning fork, six beats per second are heard. When the second tuning fork is loaded with wax and sounded with the first tuning fork, the number of beats is two per second. The natural frequency of the second tuning fork isa)334b)339c)343d)347Correct answer is option 'D'. Can you explain this answer?
Question Description
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