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Question: \( 64 \) tuning forks are arranged such that each fork produces \( 4 \) beats per second with the ne...

6464 tuning forks are arranged such that each fork produces 44 beats per second with the next one. If the frequency of the last fork is octave of the first, frequency of 16th16th fork is
A. 316 Hz316{\text{ Hz}}
B. 322 Hz{\text{322 Hz}}
C. 312 Hz{\text{312 Hz}}
D. 308 Hz{\text{308 Hz}}

Explanation

Solution

In this question, we are given each produces a beat frequency of 4 Hz{\text{4 Hz}} so the difference between the frequencies of two consecutive forks will be 44 beats per second. This forms an arithmetic progression, and we can find out the frequency of the last fork, but the last fork is the octave of the first. Using this relation, we can find out the frequency of the first fork, and using the A.P we can find out the frequency of the 16th16th fork
Beat frequency, fb=f2f1{{\text{f}}_b} = \left| {{{\text{f}}_2} - {{\text{f}}_1}} \right|
Where f1{{\text{f}}_1} and f2{{\text{f}}_2} are the frequencies of two waves.

Complete answer:
We are given,
There are 6464 tuning forks and each produces 44 beats per second
Let the frequencies of fork be f1,f2,f3, ...... , f64{{\text{f}}_{_1}},{{\text{f}}_2},{{\text{f}}_3},{\text{ }}......{\text{ , }}{{\text{f}}_{64}} respectively.
Since f2f1=f4f3=....=f64f63=4{{\text{f}}_2} - {{\text{f}}_1} = {{\text{f}}_4} - {{\text{f}}_3} = .... = {{\text{f}}_{64}} - {{\text{f}}_{63}} = 4 they form an arithmetic progression with first term as f1{{\text{f}}_{_1}} and common difference as 4{\text{4}}
Therefore, the frequency of last tuning fork, f64=f1+(n1)d{{\text{f}}_{64}} = {{\text{f}}_1} + \left( {n - 1} \right)d
f64=f1+(641)4\Rightarrow {{\text{f}}_{64}} = {{\text{f}}_1} + \left( {64 - 1} \right)4
f64=f1+252\Rightarrow {{\text{f}}_{64}} = {{\text{f}}_1} + 252
But it is given the last fork is octave of the first
Thus, f64=2f1{{\text{f}}_{64}} = 2{{\text{f}}_1}
Substituting this in the equation we get,
2f1=f1+252\Rightarrow 2{{\text{f}}_1} = {{\text{f}}_1} + 252
f1=252 Hz\Rightarrow {{\text{f}}_1} = 252{\text{ Hz}}
The Frequency of the first tuning fork is 252 Hz252{\text{ Hz}}
Now to find out the frequency of the 16th16th tuning fork we use the A.P
f16=f1+(n1)d{{\text{f}}_{16}} = {{\text{f}}_1} + \left( {n - 1} \right)d
Replacing nn with 1616 and dd with 44 we get,
f16=f1+(161)4\Rightarrow {{\text{f}}_{16}} = {{\text{f}}_1} + \left( {16 - 1} \right)4
f16=f1+60\Rightarrow {{\text{f}}_{16}} = {{\text{f}}_1} + 60
But f1=252 Hz{{\text{f}}_1} = 252{\text{ Hz}}
Therefore, f16=252+60{{\text{f}}_{16}} = 252 + 60
f16=312 Hz\Rightarrow {{\text{f}}_{16}} = 312{\text{ Hz}}
Hene, the frequency of the 16th16th fork is 312 Hz312{\text{ Hz}}
The correct option is option C.

Note:
Beats are produced when the two waves of nearby frequencies are superimposed together. This will occur when the two waves travel in the same path. Beats also cause a periodic variation in the intensity of resultant waves. If the beat frequency is greater than 10 Hz{\text{10 Hz}} then it cannot be distinguished by human ears.