Question

To tune a violin, the violinist first tunes the A string to the correct pitch of 440Hz and then bows two adjoining strings simultaneously and listens for a beat pattern. While bowing the A and E strings, the violinist hears a beat frequency increases as the tension of the E string is increased. (The E string is to be tuned to $660\;Hz$). If the tension on the E string is $80\;N$ when the beat frequency is $3\;Hz$, what tension corresponds to perfect tuning of that string?

Answer 1

Recall that the beat frequency is nothing but the difference in the frequencies of two progressive waves, here, the sound waves of the violin. Using this, we get two possible E string frequencies. See which one you can eliminate given that the beat frequency increases with an increase in tension. The frequency that suits this criterion will be the frequency of the E string that corresponds to tension of the string on perfect tuning, which is what we require to this end.

Formula Used:
$f_{beat} = |f_E-f_A|$
$f_{string} = \dfrac{1}{2L}\sqrt{\dfrac{T}{m/L}}$

Complete answer:
We know that the number of beats per second, or the beat frequency, is the difference between two frequencies. We are given that the violinist hears a beat frequency $f_{beat}$ while bowing the A and E strings with correct pitch or frequency $f_A$ and $f_E$ respectively, i.e.,
$f_{beat} = |f_E – f_A|$

On removing the modulus, we get: $\pm f_{beat} = f_E-f_A$

Given that $f_A = 440\;Hz$ and $f_{beat} = 3\;Hz$,
$\pm 3 = f_E – 440 \Rightarrow f_E = 440 \pm 3 \Rightarrow f_E = 443\;Hz$ or $437\;Hz$

Now, we are given that the violinist hears that the beat frequency increases with the E string tension.

We know that for a string vibration, the frequency of vibration of a string can be given as:
$f_{string} = \dfrac{1}{2L}\sqrt{\dfrac{T}{m/L}}$, where T is the tension of the string, m is the mass of the string and L is the length of the string.
$\Rightarrow f_{string} \propto \sqrt{T}$.

If $f_{E} = 440-3=437\;Hz$, increasing the tension would further increase this frequency, which would decrease the beat frequency. But since the beat frequency is given to increase with the string’s tension, the appropriate choice for $f_E$ would be $f_E = 440+3=443\;Hz$

At this frequency, we are given that the tension in the string is $T_{E} = 80\;N$

Using the relation $f_{string} \propto \sqrt{T_{string}}$, we can find the tension $T_{perfect}$ corresponding to the perfect tuning of this string which would produce a frequency of $f_{perfect} = 660\;Hz$ as:
$\dfrac{f_E}{f_{perfect}} = \dfrac{T_E}{T_{perfect}}$
$\Rightarrow \dfrac{443}{660} = \sqrt{\dfrac{80}{T_{perfect}}} \Rightarrow \dfrac{443^2}{660^2} = \dfrac{80}{T_{perfect}}$
$\Rightarrow T_{perfect} = \dfrac{80\times 660^2}{443^2} = 177.57\;N$

Therefore, the tension corresponding to tuning the E string to $660\;Hz$ is $\approx 178\;N$

Note:
Remember that the physical perception of frequency is commonly referred to as the pitch of a sound. A high pitch sound corresponds to a high frequency sound wave and a low pitch sound corresponds to a low frequency sound wave. Thus, for mathematical purposes the two can be treated as the same irrespective of their subjective differences.
Also do not get confused between beat frequency and wave frequency. They are different as beat frequency is obtained from the superposition of individual waves while wave frequency is an innate property of the wave produced.