Question

Two guitarists attempt to play the same note of wavelength 64.8 cm at the same time, but one of the instruments is slightly out of tune and plays a note of wavelength 65.2 cm instead. What is the frequency of the beats these musicians hear when they play together? Should the out of tune guitarists tighten or loosen his guitar string to correctly tune his guitar?

Answer 1

To solve this problem it is necessary to apply the concepts related to frequency, wavelength and speed of sound.

The frequency is defined as,

$f=\frac{v}{\lambda}$

Where,

$v = 340m/s = 34000cm/s \rightarrow$ Speed of sound

$\lambda = Wavelength$

Calculating the values of the frequencies with the two wavelengths we have

$f_1 = \frac{v}{\lambda_1}$

$f_1 = \frac{34000}{64.8}$

$f_1=524.69Hz$

And,

$f_1 = \frac{v}{\lambda_2}$

$f_2 = \frac{34000}{65.2}$

$f_2=521.47Hz$

Therefore the frequency of the beat is

$f_{beat} = f_1-f_2$

$f_{beat} = 524.69-521.47$

$f_{beat} = 3.22Hz$

Finally we can get the wavelenght through the same equation, then

$\lambda = \frac{v}{f}$

$\lambda = \frac{34000}{3.22}$

$\lambda = 10.559cm$

We can conclude that the frequency of the beats these musicians hear when they play together is 3.22 Hz and the out of tune guitarists should tighten his guitar string to correctly tune it.