Question

The frequency of tuning forks $A$ and $B$ are respectively $3\%$ more and $2\%$ 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
$A)\text{ }98$
$B)\text{ 100}$
$C)\text{ 103}$
$D)\text{ 105}$

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:
$f=\left| {{f}_{1}}-{{f}_{2}} \right|$

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 $3\%$ more than the frequency of tuning fork $C$.
$\therefore {{f}_{A}}={{f}_{C}}+\dfrac{3}{100}{{f}_{C}}=\dfrac{100+3}{100}{{f}_{C}}=\dfrac{103}{100}{{f}_{C}}$ --(1)
Also, according to the question, the frequency of tuning fork $B$ is $2\%$ less than the frequency of tuning fork $C$.
$\therefore {{f}_{B}}={{f}_{C}}-\dfrac{2}{100}{{f}_{C}}=\dfrac{100-2}{100}{{f}_{C}}=\dfrac{98}{100}{{f}_{C}}$ --(2)
According to the question, the frequency of the beats produced when tuning forks $A$ and $B$ are struck simultaneously is $f=5\text{ /sec}=5Hz$ $\left( \because 1\text{ se}{{\text{c}}^{-1}}=1Hz \right)$
The beats frequency $f$ produced when two waves of nearby frequencies ${{f}_{1}}$ and ${{f}_{2}}$ superimpose is given by
$f=\left| {{f}_{1}}-{{f}_{2}} \right|$ --(3)
Now, using (3), we get
$f=\left| {{f}_{A}}-{{f}_{B}} \right|$
Now using (1) and (2), we get
$f=\left| \dfrac{103}{100}{{f}_{C}}-\dfrac{98}{100}{{f}_{C}} \right|=\left| \dfrac{103-98}{100}{{f}_{C}} \right|=\left| \dfrac{5}{100}{{f}_{C}} \right|=\dfrac{5}{100}{{f}_{C}}$ $\left( \because {{f}_{C}}>0 \right)$
$\therefore 5=\dfrac{5}{100}{{f}_{C}}$
$\therefore {{f}_{C}}=100Hz$ --(4)
Putting (4) in (1), we get
${{f}_{A}}=\dfrac{103}{100}{{f}_{C}}=\dfrac{103}{100}\times 100=103Hz$
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.