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.

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

the second one and the forth one are in the purple because they are opinions and the rest are in the yellow

explanation:

answer; how far out of place which the object finished relative to the starting location;