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 $660Hz$.)
If the tension on the $E$ string is $80.0N$ when the beat frequency is $3Hz$, what tension corresponds to perfect tuning of that string ?

Answer 1

$f_{beat}=|f_E-f_A|$

removing the modulus we get,

$\pm f_{beat}=f_E-f_A$

given: $f_A=440Hz$ and $f_{beat}=3$

we get,

$\pm 3=f_E-440\Rightarrow f_E=443Hz,437Hz$

we also know, $f\propto \sqrt{T}$

So if 437 is the frequency then on increasing the tension it may go to 438, then the beat decreases.

But it is given that the beat increases with tension.

So,

correct answer $f_E=443Hz$

therefore, $\frac{f_E}{f_{final}}=\sqrt{\frac{T_E}{T_{final}}}$

$\Rightarrow \frac{443}{600}=\sqrt{\frac{80}{T_{final}}}$

$T_{final}=147N$