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Speed of a Wave in Steel Wire

  • Thread starter gmmstr827
  • Start date
  • #1
86
1

Homework Statement



A 5.0-kg ball hangs from a steel wire 1.00 mm in diameter and 5.00 m long. What would be the speed of a wave in the steel wire?
Hint: Density of steel = þ_steel = 7.8 * 10^3 kg/m^3
Assume that the string is a cylinder to calculate the volume.

m = 5.0 kg
l (length) = 5.00 m
d = 1.00 mm = 0.001 m
r = 0.0005 m
þ_steel = 7.8 * 10^3 kg/m^3
g = 9.8 m/s^2

Homework Equations



V(volume)_cylinder = πlr^2
m = þV
F = mg
µ = m/l
v(velocity) = √(F/µ)

The Attempt at a Solution



V_cylinder = πlr^2 = π (5.00 m) (0.0005m)^2 ≈ 4.0 * 10^-6 m^3
m = þV = (7.8 * 10^3 kg/m^3) (4.0 * 10^-6 m^3) ≈ 0.0312 kg
F = mg = (0.0312 kg) (9.8 m/s^2) ≈ 0.306 N
µ = m/l = (5.0 kg)/(5.00 m) = 1 kg/m
v = √(F/µ) = √((0.306 N)/(1 kg/m)) = √(0.306) m/s ≈ 0.553 m/s

So, the speed of the wave in the steel wire would be approximately 0.553 m/s.

^^^ Does all of that look correct?
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31

The Attempt at a Solution



V_cylinder = πlr^2 = π (5.00 m) (0.0005m)^2 ≈ 4.0 * 10^-6 m^3
m = þV = (7.8 * 10^3 kg/m^3) (4.0 * 10^-6 m^3) ≈ 0.0312 kg
F = mg = (0.0312 kg) (9.8 m/s^2) ≈ 0.306 N
µ = m/l = (5.0 kg)/(5.00 m) = 1 kg/m
v = √(F/µ) = √((0.306 N)/(1 kg/m)) = √(0.306) m/s ≈ 0.553 m/s

So, the speed of the wave in the steel wire would be approximately 0.553 m/s.

^^^ Does all of that look correct?
µ would be the mass per unit length of the wire, so you need to redo that part.

Also the F=mg is the force produced due to the 5 kg mass.
 
  • #3
86
1
µ would be the mass per unit length of the wire, so you need to redo that part.

Also the F=mg is the force produced due to the 5 kg mass.
Thanks for the help!

Ok, so now I have:

F = mg = (5.0 kg) (9.8 m/s^2) = 49 N
µ = m/l = (0.0312 kg)/(5.00 m) ≈ 0.00624 kg/m
v = √(F/µ) = √((49 N)/(0.00624 kg/m)) ≈ 88.615 m/s

Is THAT correct?
 
  • #4
rock.freak667
Homework Helper
6,230
31
That looks more correct.
 

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