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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 340Hz340{\text{Hz}}. 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.

Explanation

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=f1f2{f_{beat}} = {f_1} - {f_2} where f1{f_1} and f2{f_2} 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{f_D} = 340{\text{Hz}}
Let fC{f_C} be the unknown frequency of the tuning fork C.
Step 2: Using the expression for beat frequency, we can find the frequency fC{f_C} before and after filing.
The frequency of the beats produced is given by, fbeat=fDfC{f_{beat}} = {f_D} - {f_C} ------- (1)
From equation (1), we can express the frequency of fork C as fC=fD+fbeat{f_C} = {f_D} + {f_{beat}} .
But we know the frequency of a sound wave can be fC=fD+fbeat{f_C} = {f_D} + {f_{beat}} -------- (2)
or fC=fDfbeat{f_C} = {f_D} - {f_{beat}} ------- (3).

Before filing
We consider the case where no filing is added to the tuning fork C.
Substituting for fD=340Hz{f_D} = 340{\text{Hz}} and fbeat=8{f_{beat}} = 8 in equations (2) and (3) we get, fC=340+8=348Hz{f_C} = 340 + 8 = 348{\text{Hz}} or fC=3408=332Hz{f_C} = 340 - 8 = 332{\text{Hz}}
So the two possible frequencies of fork C before filing are fC=348Hz{f_C} = 348{\text{Hz}} or fC=332Hz{f_C} = 332{\text{Hz}}.

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{f_C} of the fork C increases.
Now if fC=348Hz{f_C} = 348{\text{Hz}}, then on filing, it would become fC>348Hz{f_C} > 348{\text{Hz}}. Then the relation for the beat frequency fbeat=fC340=4{f_{beat}} = {f_C} - 340 = 4 suggests that fC=344Hz{f_C} = 344{\text{Hz}}, but this is impossible as 344Hz<348Hz344{\text{Hz}} < 348{\text{Hz}}. So, the frequency of the tuning fork C will not be 348Hz348{\text{Hz}} i.e., fC348Hz{f_C} \ne 348{\text{Hz}} .
Thus the frequency of fork C before filing is fC=332Hz{f_C} = 332{\text{Hz}} .

Note: When we consider fC=332Hz{f_C} = 332{\text{Hz}}, we know that after filing the frequency of the tuning fork C must increase i.e., fC<332Hz{f_C} < 332{\text{Hz}}. Then the relation of beat frequency, fbeat=340fC=4{f_{beat}} = 340 - {f_C} = 4 implies that fC=336Hz{f_C} = 336{\text{Hz}} i.e., the frequency of the fork C has increased from 332Hz332{\text{Hz}} to 336Hz336{\text{Hz}} on filing. While we calculate the beat frequency, we subtract the smaller frequency from the bigger one so that the beat frequency remains positive.