Question

two tuning forks A and B produce 10 beats per second. on loading a small ring on one prong of B again 10 beats per second are produced.what was the frequency of B before loading small ring if now frequency of B is 430Hz? give reason for your answer

Answer 1

When we consider a system of forks, the frequency of the beats can be found as

$f_{beat}=|f_A-f_B|$

Then, we have to consider that when the mass increases the frequency decreases thus $f_B>f'_B$

$f_{beat_1}=|f_A-f_B|=10\,Hz \\ f_{beat_2}=|f_A-f'_B|=10\,Hz$

Then, since we know that $f'_B=430\,Hz$, we use the second equation for the beat frequency and we consider that $f'_B<f_A$ to find the value of f_A and then determine f_B:

$|f_A-f'_B|=10\,Hz \\ \to (f_A-f'_B)=10\,Hz \\ \implies f_A=f'_B+10\,Hz=440 Hz$

With the last result, this will only work if we consider that $f_B>f_A$ to solve the last equation (otherwise we would get the same frequency for f_B and f'_B and this would not be consistent with the physical phenomena of the increase of frequency as a result of adding mass to an oscillating system):

$|f_A-f_B|=10\,Hz \\ \to -(f_A-f_B)=10\,Hz \\ \implies f_B=f_A+10\,Hz=450 Hz \\ \text{the last result also satisfies the inequality } f_B>f'_B$

In conclusion, we find that f_B = 450 Hz.

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.