two identical guitar strings are stretched with the same ten

two identical guitar strings are stretched with the same tension between two supports that are not the same distance apart. the fundamental frequency of the higher pitched string is 380Hz and the speed of the transverse waves in both wires is 200m/s. how much longer is the lower pitched string if the bear frequency is 3 Hz? in m

Solution

Fundamental frequency f = v / 2L

Where f = 380 Hz

           v = 200 m/s

From this length of higher pitched string L = v / 2f

                                                             = 200 /(2 x380)

                                                             = 0.2631 m

Frequency of lower pitched string f \' = 380 -3 = 377 Hz

f \' / f = L / L \'

from this L \' = L ( f/f \' )

                 = 0.2631 (380/377)

                 = 0.2652 m