Asked by James
The tension in a 2.6-m-long, 1.1-cm-diameter steel cable (ρ = 7800 kg/m3) is 860 N. What is the fundamental frequency of vibration of the cable?
The answer is 6.55 Hz.
I know that I'm going to use the formula:
f=nsqrt(T/4L^2u)
However, I'm not sure how to get the u for this equation. Any help is appreciated.
The answer is 6.55 Hz.
I know that I'm going to use the formula:
f=nsqrt(T/4L^2u)
However, I'm not sure how to get the u for this equation. Any help is appreciated.
Answers
Answered by
Damon
You need to find the mass of the cable and divide it by the length to get the mass per unit length.
Volume of cable = pi r^2 L
Mass of cable = pi r^2 L * 2.6
so
mass/length = 7800 * pi * (.011)^2
Volume of cable = pi r^2 L
Mass of cable = pi r^2 L * 2.6
so
mass/length = 7800 * pi * (.011)^2
Answered by
James
Thank you! I was able to figure out the correct answer. However, in this case shouldn't (.011^2) be (.011/2)^2?