Question

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.

Answer 1

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

$v=\sqrt{F/U}$

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

$v=\sqrt{F/U}$

$v2=\sqrt1.01F/U}$

v2=$\sqrt{1.01} v$

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