Question

Two identical mandolin strings under 200 N of tension are sounding tones with frequencies of 590 Hz. The peg of one string slips slightly, and the tension in it drops to 192 N. How many beats per second are heard?

Answer 1

To solve this problem it is necessary to apply the concepts related to frequency and vibration of strings. Mathematically the frequency can be expressed as

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

Then the relation between two different frequencies with same wavelength would be

$\frac{f'}{f} = \frac{v'/\wavelength}{v/\wavelength}$

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

The beat frequency heard when the two strings are sounded simultaneously is

$f_{beat} = f-f'$

$f_{beat} = f(1-\frac{f'}{f})$

$f_{beat} = f(1-\frac{v'}{v})$

We have the velocity of the transverse waves in stretched string as

$v = \sqrt{\frac{T}{\mu}}$

$v = \sqrt{\frac{200N}{\mu}}$

And,

$v' = \sqrt{\frac{196N}{\mu}}$

Therefore the relation between the two is,

$\frac{v'}{v} = \sqrt{\frac{192}{200}}$

$\frac{v'}{v} = \sqrt{0.96}$

Finally substituting this value at the frequency beat equation we have

$f_{beat} = 590(1-\sqrt{0-96})$

$f_{beat} = 11.92Hz$

Therefore the beats per second are 11.92Hz