Two tuning forks P and Q give 4 beats per second. On loading Q lightly with wax, we get 3 beats per second. What is the frequency of Q before and after loading it if the frequency of P is 512 Hz.
Solution
Beat frequency is the difference of frequencies of the two sounding sources and given by, f = f2 – f1.
Given Frequency of fork P = 512 Hz
No of beats when Q is not waxed = 4
No of beats when Q is waxed = 3
Required (i) Frequency of Q before waxing it
(ii) Frequency of Q when it is waxed.
Formula f = f2 – f1
Case 1, when the second fork is not waxed.
4 = f2 – 512 OR f2 = 4 + 512 = 516 Hz
Case 2, when the second fork is waxed.
3 = f2 – 512 OR f2 = 3+512 = 515 Hz
Hello plz ive a confusion. In our book f1-f2 formula is used while you used f2-f1??
Omaima
It does not make any difference if you take f2 – f1 or f1 – f2. The actual sense of the formula is to find the difference between the two frequencies. However, it is more appropriate to take f2 – f1 if f2 > f1 or f1 – f2 if f1 > f2. Though in both cases, the absolute value of the answer will be same.
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