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
[tex]v=\sqrt{F/U}[/tex]
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
[tex]v=\sqrt{F/U}[/tex]
[tex]v2=\sqrt1.01F/U}[/tex]
v2=[tex]\sqrt{1.01} v[/tex]
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
