The G string on a violin has a fundamental frequency of 196 Hz. It is 30.0 cm long. While this string is sounding, a nearby violinist effectively shortens (by sliding her finger down the string) the G string on her identical violin until a beat frequency of 3.5 Hz is heard between the two strings. When this occurs, how far down the string did she slide her finger

L = 29.47 cm

Explanation:

wavelength = twice the string length = (wave velocity on string)/(frequency)

Thus;

2(L_o) = v/(f_o)

v = 2(L_o)(f_o) - - - eq1

wave velocity on string = √[(tension)/(mass per unit length)]

If we assume that the 2 violins have matching string gauges and they're tuned exactly the same, and the necks of the violins are the same length, the tensions needed to get the same frequency on open G strings would be the same. Hence, we can deduce that the wave velocities on the 2 strings would also be the same.

The note on one violin has to differ from the note on the other violin by the beat frequency. A shorter string produces a higher pitch (higher frequency). Thus,

v = 2L(f_o + f_b) - - - - eq2

Equating equation 1 and 2 to obtain;

2(L_o)(f_o) = 2L(f_o + f_b)

(Lo)(fo) = L(fo + fb)

(30 cm)(196 Hz) = (L)(196Hz + 3.5Hz)

5880 = 199.5L

L = 5880/199.5

L = 29.47 cm

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