Question

A string is stretched between fixed points separated by $75cm$. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is
A. $10.5Hz$
B. $105Hz$
C. $1.05Hz$
D. $1050Hz$

Answer 1

To solve the given question use the formula for the ${{n}^{th}}$ resonant frequency for a string with both of its ends fixed. Then use the given data that there are no other resonant frequencies between the given frequencies.

Formula used:
$f=\dfrac{nV}{2L}$
where $n$ is a natural number, $V$ is the speed of the wave and $L$ is the length of the string.

Complete step by step answer:
A resonant frequency of a string is its natural frequency of a standing wave that is produced in it. The standing waves in the string are called harmonics.A string of a fixed length can have many standing waves. The frequencies of these standing waves are called resonant frequencies. The fundamental frequency is the least possible resonant frequency. The frequency of each harmonic is different.
When both the ends of a string are fixed, the resonant frequency of a standing wave is given as $f=\dfrac{nV}{2L}$ ….. (i),
where n is a natural number, V is the speed of the wave and L is the length of the string.
These frequencies are named as fundamental frequency, first harmonics, second harmonic and so on, depending on the value of n.

Let the value of n for the given two resonant frequencies (420 Hz and 315 Hz) be ${{n}_{1}}$ and ${{n}_{2}}$ respectively.For a fixed tension in the string, the speed of the waves remains the same. Therefore, the speed of both the waves is V.
By substituting the values in (i) we get
$420=\dfrac{{{n}_{1}}V}{2L}$ …… (ii)
And $315=\dfrac{{{n}_{2}}V}{2L}$ ….. (iii).
Now, divide (ii) by (iii).
$\Rightarrow \dfrac{420}{315}=\dfrac{\dfrac{{{n}_{1}}V}{2L}}{\dfrac{{{n}_{2}}V}{2L}}=\dfrac{{{n}_{1}}}{{{n}_{2}}}$ ….. (iv).

It is also given that there are no other resonant frequencies for the given string. This means that the resonant frequencies are consecutive.
Therefore, $\Rightarrow {{n}_{1}}={{n}_{2}}+1$.
Substitute these values in (iv).
$\Rightarrow \dfrac{420}{315}=\dfrac{{{n}_{2}}+1}{{{n}_{2}}}$
$\Rightarrow 420{{n}_{2}}=315{{n}_{2}}+315$
$\Rightarrow {{n}_{2}}=\dfrac{315}{105}=3$.
Also, $\Rightarrow {{n}_{1}}={{n}_{2}}+1=3+1=4$.

The least possible resonant frequency (fundamental frequency) is when $n=1$. Therefore, the value of the fundamental frequency is ${{f}_{1}}=\dfrac{V}{2L}$.
But we know that $420=\dfrac{{{n}_{1}}V}{2L}$.
$\Rightarrow 420={{n}_{1}}{{f}_{1}}$.
And ${{n}_{1}}=4$
$\Rightarrow 420=4{{f}_{1}}$
$\therefore {{f}_{1}}=105Hz$
Therefore, the lowest possible resonance frequency for the given string is $105 Hz$.

Hence, the correct option is B.

Note: From the above solution we can observe that the difference of two consecutive resonant frequencies of a string with both end fixed is equal to ${{f}_{n+1}}-{{f}_{n}}=\dfrac{V}{2L}$.
And this means that ${{f}_{n+1}}-{{f}_{n}}={{f}_{1}}$,
where $f$ is the fundamental frequency.
If we knew this then we could have easily found the answer by just subtracting the two given frequencies.