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The frequency of tuning fork A is 2% more than the frequency of a standard tuning fork. Fre-quency of tuning fork B is 3%. less than the frequency of standard fork. If 6 beats per second areheard when two forks Aand B are excited. The frequency of tuning fork A is

Question

The frequency of tuning fork A is 2% more than the frequency of a standard tuning fork. Fre-quency of tuning fork B is 3%. less than the frequency of standard fork. If 6 beats per second areheard when two forks Aand B are excited. The frequency of tuning fork A is

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Solution

Let's denote the frequency of the standard tuning fork as f.

  1. According to the problem, the frequency of tuning fork A is 2% more than the standard frequency. So, the frequency of A (f_A) can be calculated as follows: f_A = f + 0.02f = 1.02f

  2. Similarly, the frequency of tuning fork B is 3% less than the standard frequency. So, the frequency of B (f_B) can be calculated as follows: f_B = f - 0.03f = 0.97f

  3. The problem states that 6 beats per second are heard when tuning forks A and B are excited. The beat frequency is the absolute difference between the frequencies of the two tuning forks. So, we can set up the following equation: |f_A - f_B| = 6

  4. Substituting the expressions for f_A and f_B from steps 1 and 2 into the equation from step 3, we get: |1.02f - 0.97f| = 6 |0.05f| = 6

  5. Since frequency cannot be negative, we can remove the absolute value, giving us: 0.05f = 6

  6. Solving for f, we get: f = 6 / 0.05 = 120 Hz

  7. Therefore, the frequency of tuning fork A is: f_A = 1.02f = 1.02 * 120 = 122.4 Hz

So, the frequency of tuning fork A is 122.4 Hz.

This problem has been solved

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