Question
Question: Two tuning forks A and B produce notes of frequencies \(258{\text{Hz}}\) and \(262{\text{Hz}}\) . An...
Two tuning forks A and B produce notes of frequencies 258Hz and 262Hz . An unknown note sounded with certain beats. When the same note is sounded with B, the beat frequency gets doubled. Find the unknown frequency.
A) 250Hz
B) 252Hz
C) 254Hz
D) 256Hz
Solution
When two different frequencies are sounded together beats are produced having a frequency equal to the difference between the two frequencies. We can obtain two equations corresponding to the frequency of the beat produced when the unknown frequency was sounded with tuning forks A and B respectively. Solving these two equations will help us to determine the unknown frequency.
Formula used:
-The beat frequency is given by, fbeat=±(f1−f2) where f1 and f2 are the frequencies sounded together.
Complete step by step solution.
Step 1: List the given frequencies of the two tuning forks.
The frequency of tuning fork A is given to be fA=258Hz .
The frequency of tuning fork B is given to be fB=262Hz .
Let f and fbeat be the unknown frequency and the frequency of beats produced when A was sounded with the unknown note respectively.
Step 2: Express the beat frequency when each tuning fork was sounded with the unknown frequency to determine the beat frequency fbeat .
The frequency of beats produced when fork A was sounded with the unknown note can be expressed as fbeat=±(fA−f) ------- (1)
Since the frequency of beats produced when fork B was sounded with the unknown note is said to be doubled, we have 2fbeat=±(fB−f) ------- (2)
We assume the differences in both equations to be positive and subtract equation (1) from (2) to get, 2fbeat−fbeat=(fB−f)−(fA−f)
⇒fbeat=fB−fA -------- (3)
Substituting for fA=258Hz and fB=262Hz in equation (3) we get, fbeat=262−258=4Hz
Thus the beat frequency is obtained to be fbeat=4Hz .
Step 3: Express equation (1) in terms of the unknown frequency and make necessary substitutions.
From equation (1) we can express the unknown frequency as f=+(fA−fbeat) ------- (4)
Substituting for fA=258Hz and fbeat=4Hz in equation (4) we get, f=∣258−4∣=254Hz
∴ the unknown frequency is obtained to be f=254Hz .
Hence the correct option is C.
Note: Here it is mentioned that the beat frequency of tuning fork B is the double of the beat produced by tuning fork A. This suggests that the unknown frequency f is less than the given frequency of fork B fB so that a greater difference in frequency (which is the beat frequency) is maintained for fork B. In the above calculations, we only considered the positive value fbeat=+(fA−f) . If we were to consider the negative value i.e., fbeat=−(fA−f)=(f−fA) , then we obtain f=fA+fbeat=258+4=262Hz . But since f cannot be greater than fB we discard the negative value.