Question

Please someone help me I'm desperate!!!!

Two steel guitar strings have the same length. String A has a diameter of 0.59 mm and is under 420.0 N of tension. String B has a diameter of 1.3 mm and is under a tension of 840.0N .

Find the ratio of the wave speeds,VA/VB , in these two strings.

Answer 1

Wave speed is proportional to
[Tension/(linear density)]^1/2

Linear density is the density per unit string length. String A has 1/2 the tension and (0.59/1.3)^2 = 0.206 times the linear density of String B.

The wave speed ratio is therefore
VA/VB = sqrt [(1/2)/0.206] = 1.558

The length does not matter for the wave speed, but does help determine the resonant standing wave frequencies.

Answer 2

wave speed= sqrt (tension/(mass/length))

now, mass/length in the strings (assuming same material) is proportional to area, or diamter^2

wavespeed=Constant*sqrt (tension/diameter^2)

so
wavespeed1/wavespeed2= sqrt(Tension1/tension2 * diameter2^2/diameter1^2)

Answer 3

Very close. Remember to change mm to meters. Greatly effects answer.

Va/Vb = sqrt[(Tension1/Tension2)*(Diameter2^2/Diameter1^2)]

Va/Vb = sqrt[(420/840)*((5.9E-4)^2/(1.3E-3)^2)]