Q41E - University Physics with Modern Physics

Question

A violinist is tuning her instrument to concert A $\left(\mathbf{440}\mathbf{Hz}\right)$. She plays the note while listening to an electronically generated tone of exactly that frequency and hears a beat frequency of $\mathbf{3}\mathbf{Hz}$, which increases to $\mathbf{4}\mathbf{Hz}$when she tightens her violin string slightly. (a) What was the frequency of the note played by her violin when she heard the $\mathbf{3}\mathbf{Hz}$ beats? (b) To get her violin perfectly tuned to concert A, should she tighten or loosen her string from what it was when she heard the $\mathbf{3}\mathbf{Hz}$beats?

Step-by-Step Solution

Answer

(a) The frequency of the violin is $f_v=443\text{\hspace{0.17em}Hz}$

(b) For perfect tuning, the violinist should loosen the string.

Step 1: Beat frequency formula

The beat frequency is the difference in frequency of the superimposed waves ${\mathbf{f}}_{\mathrm{beat}}\mathbf{=}|{\mathbf{f}}_1\mathbf{-}{\mathbf{f}}_2|$ .

Since, we have violin and electrically generate frequency, we will have

$\begin{array}{l}{f}_{beat}=f_v-f_e\\ f_v=f_{beat}+f_e\end{array}$ 

Step 2: Calculate the violin frequency

Given that $f_{beat}=3\text{\hspace{0.17em}Hz}$ and $f_e=440\text{\hspace{0.17em}Hz}$ .

$\begin{array}{l}{f}_v=f_{beat}+f_e\\ f_v=3\text{\hspace{0.17em}Hz}+440\text{\hspace{0.17em}Hz}\\ f_v=443\text{\hspace{0.17em}Hz}\end{array}$

Step 3: Make beat frequency equal to zero

In order to get her violin perfectly tuned, the beat frequency must be zero. This means that  $f_v$ should decrease by  $3\text{\hspace{0.17em}Hz}$. To decrease the frequency, the violinist should loosen the string.