Answer:
a,V=311.15m/s
b.246Hz
c.245Hz
d. 1.4m
Explanation:
One of the 63.5-cm-long strings of an ordinaryguitar is tuned to produce the note {\rm B_3}(frequency 245 Hz) when vibrating inits fundamental mode.
A.Find the speed of transverse waves on this string.B.If the tension in this string is increased by1.0%, what will be the new fundamental frequency ofthe string?C.If the speed of sound in the surrounding air is 344m/s, find the frequency of the sound wave produced in theair by the vibration of the {\rm B_3} string.D.If the speed of sound in the surrounding air is 344m/s, find the wavelength of the sound wave produced in theair by the vibration of the {\rm B_3} string
NOTE
speed is distance by the wave per time
frequency is the number of oscillation the wave front makes in one seconds
the wave speed is given by
recall also that the wave speed is v=f lambda
for a standing wave , we know the fundamental frequency of a string is
f1=v/2L
L=length of the string
f1=245Hz
V=?
L=0.635m
V=245*2*0.635
V=311.15m/s
b. tension in the string is increased by 1%
F2=F+1%
f2=101F%
substituting for F2 into this equation
v2=
v2=1.01^0.5*311.15m/s
v2=312.7m/s
for the new fundamental frequency we have
f2=312.7/2*0.635
f2=246Hz
c. the frequency of the sound wave equal the frequency of the string that created it
c. fs=245Hz
d. speed of sound in air344m/s
v=344m/s
v=f*lambda
lambda is the wavelength
344=245*lambda
344/245=1.40m
wavelength of string B3 is 1.4m
