Question
Question: A tuning fork C produces 8 beats per second with another tuning fork D of frequency \(340{\text{Hz}}...
A tuning fork C produces 8 beats per second with another tuning fork D of frequency 340Hz. When the prongs of the tuning fork C are filed a little, the number of beats produced per second decreases to 4. Find the frequency of the tuning fork C before filing its prongs.
Solution
When the prongs of a tuning fork are filed, the frequency of that tuning fork will increase. Here, it is mentioned that the beats produced per second decreases when tuning fork C is filed. The beats produced per second refer to the difference between the frequencies of the two tuning forks.
Formula used:
-The frequency of the beats produced is given by, fbeat=f1−f2 where f1 and f2 are the frequencies of the two sound waves created by two tuning forks.
Complete step by step answer.
Step 1: List the parameters known from the question.
The beat frequency when the forks C and D were sounded together is given to be 8 beats per second.
As the tuning fork C of unknown frequency is filed, the beat frequency becomes 4 beats per second.
The frequency of the tuning fork D is given to be fD=340Hz
Let fC be the unknown frequency of the tuning fork C.
Step 2: Using the expression for beat frequency, we can find the frequency fC before and after filing.
The frequency of the beats produced is given by, fbeat=fD−fC ------- (1)
From equation (1), we can express the frequency of fork C as fC=fD+fbeat .
But we know the frequency of a sound wave can be fC=fD+fbeat -------- (2)
or fC=fD−fbeat ------- (3).
Before filing
We consider the case where no filing is added to the tuning fork C.
Substituting for fD=340Hz and fbeat=8 in equations (2) and (3) we get, fC=340+8=348Hz or fC=340−8=332Hz
So the two possible frequencies of fork C before filing are fC=348Hz or fC=332Hz.
After filing
We now consider the case where filing is added to the tuning fork C. The number of beats produced becomes 4. And the frequency fC of the fork C increases.
Now if fC=348Hz, then on filing, it would become fC>348Hz. Then the relation for the beat frequency fbeat=fC−340=4 suggests that fC=344Hz, but this is impossible as 344Hz<348Hz. So, the frequency of the tuning fork C will not be 348Hz i.e., fC=348Hz .
Thus the frequency of fork C before filing is fC=332Hz .
Note: When we consider fC=332Hz, we know that after filing the frequency of the tuning fork C must increase i.e., fC<332Hz. Then the relation of beat frequency, fbeat=340−fC=4 implies that fC=336Hz i.e., the frequency of the fork C has increased from 332Hz to 336Hz on filing. While we calculate the beat frequency, we subtract the smaller frequency from the bigger one so that the beat frequency remains positive.