Question

Two identical guitar strings are stretched with the same tension between supports that are not the same distance apart. The fundamental frequency of the higher-pitched string is 360Hz, and the speed of transverse waves in both wires is 150 m/s. How much longer is the lower-pitched string if the beat frequency is 4Hz?

Answer 1

given

The fundamental frequency of the higher-pitched string is 360 Hz

now n = 360 - 4 = 356 Hz

the speed of transverse wave is 150 m/sec = V

the lower-pitched string if the beat frequency is 4Hz

the fundamental frequency is n = V / 2 L

so the relation is n_low / n_high = L_high / L_low

356 / 360 = L / L'

L ' = 1.011 L

from here the lower pitched string is having 1.011 times of that of the higher

again using n = V / 2 L

L = V / 2 n

L = 150 / 2 X 360

L = 0.2083 m

then L ' = 1.011 L

L ' = 1.011 X 0.2083

L ' = 0.21059 m