Question

The frequency of tuning forks $A$ and $B$ are respectively $3\mathrm{\%}$ more and $2\mathrm{\%}$ less than the frequency of tuning fork $C$. When $A$ and $B$ are simultaneously excited, $5\text{ beats/sec}$ are produced. Then the frequency of the tuning fork $A\left(\text{In Hz}\right)$ is
$A)98$
$B)\text{ 100}$
$C)\text{ 103}$
$D)\text{ 105}$

Answer 1

Hint: This problem can be solved by using the formula for the beats produced by two nearby frequencies. We can write the frequencies of tuning forks A and B in terms of that of C and then find the frequency of C from which we can further get the frequency of A.

Formula used:
$f=|f_1-f_2|$

Complete step by step answer:
We will write the frequency of the tuning forks of A and B in terms of C and then find C from the beat frequency.
Let the frequency of tuning fork $A$ be $f_A$.
Let the frequency of tuning fork $B$ be $f_B$.
Let the frequency of the tuning fork $C$ be $f_C$.
According to the question, the frequency of tuning fork $A$ is $3\mathrm{\%}$ more than the frequency of tuning fork $C$.
$\therefore f_A=f_C+{\displaystyle \frac{3}{100}}{f}_C={\displaystyle \frac{100+3}{100}}{f}_C={\displaystyle \frac{103}{100}}{f}_C$ --(1)
Also, according to the question, the frequency of tuning fork $B$ is $2\mathrm{\%}$ less than the frequency of tuning fork $C$.
$\therefore f_B=f_C-{\displaystyle \frac{2}{100}}{f}_C={\displaystyle \frac{100-2}{100}}{f}_C={\displaystyle \frac{98}{100}}{f}_C$ --(2)
According to the question, the frequency of the beats produced when tuning forks $A$ and $B$ are struck simultaneously is $f=5\text{ /sec}=5Hz$ $(\because 1\text{ se}{\text{c}}^{-1}=1Hz)$
The beats frequency $f$ produced when two waves of nearby frequencies $f_1$ and $f_2$ superimpose is given by
$f=|f_1-f_2|$ --(3)
Now, using (3), we get
$f=|f_A-f_B|$
Now using (1) and (2), we get
$f=|{\displaystyle \frac{103}{100}}{f}_C-{\displaystyle \frac{98}{100}}{f}_C|=\left|{\displaystyle \frac{103-98}{100}}{f}_C\right|=\left|{\displaystyle \frac{5}{100}}{f}_C\right|={\displaystyle \frac{5}{100}}{f}_C$ $(\because f_C>0)$
$\therefore 5={\displaystyle \frac{5}{100}}{f}_C$
$\therefore f_C=100Hz$ --(4)
Putting (4) in (1), we get
$f_A={\displaystyle \frac{103}{100}}{f}_C={\displaystyle \frac{103}{100}}\times 100=103Hz$
Hence, the frequency of tuning fork $A$ is $103Hz$.
Hence, the correct option is $C)\text{ 103}$.

Note: Students could have made the mistake in understanding the language and determining the relationship of the frequencies of tuning forks A and B with that of C. A very common mistake is that students think that the frequency of A and B are the respective percentages of the frequency of C and cannot catch that A is more than C by that percentage and B is less than C by that percentage. This will lead to a completely wrong result on the part of the student.