Calculating Speed of Sound in Nickel using Atomic Properties

[jhyrman]

Homework Statement

From WebAssign:
One mole of nickel (6.02e23 atoms) has a mass of 59 g, and its density is 8.9 g/cm3. You have a bar of nickel 2.44 m long, with a square cross section, 1.8 mm on a side. You hang the rod vertically and attach a 45 kg mass to the bottom, and you observe that the bar becomes 1.6 mm longer. Next you remove the 45 kg mass, place the rod horizontally, and strike one end with a hammer. How much time T will elapse before a microphone at the other end of the bar will detect a disturbance? (Assume a simple cubic lattice for nickel.)

Homework Equations

Young's modulus, diameter of an atom, v=ωd

The Attempt at a Solution

tl;dr - I'm getting 0.000538 seconds, not the right answer.

Atomic Size and Mass:

1) convert given density to kg/m^3 = 8900kg/m^3
2) convert to moles/m^3 (kg/m^3 * mol/kg) = 150847 mol/m^3 (not rounding in my actual calculations)
3) convert to atoms/m^3 (6.022^23 atoms/mol) = 9.084e28 atoms/m^3
4) take the cube root to get the number of atoms per meter, = 4495309334 atoms/m
5) take the reciprocal to get the diameter of an atom, = 2.2245e-10 m/atom
6) find the mass of one atom (kg/mol * mol/atoms) = 9.7974e-26 kg/atom

Young's Modulus: Y=(F/A)/(dL/L)

1) F=mg = (45kg)(9.8N/kg) = 441 N
2) A = (0.0018m)^2 = 3.5344e-6 m^2
3) dL = 0.0016m
4) L = 2.44m
5) Y = 1.834e11 N/m^2

Interatomic Spring Stiffness: Ks,i = dY

1) From above, diameter of one atom = 2.2245e-10 m
2) From above, Y = 1.834e11 N/m^2
3) Ks,i = 40.799 N/m (not rounding in my actual calculations)

Speed of Sound: v = ωd

1) ω = √(Ks,i / m,a)
2) From above, Ks,i = 40.799 N/m
3) From above, m,a = 9.7974e-26 kg
4) ω=2.0406e13 N/m*kg
5) From above, d=2.2245e-10 m
6) v=ωd = 4539 m/s (not rounding in actual calculations)

Time Elapsed:

1) length sound traveled = L+dL = 2.44166 m
2) From above, speed of sound = 4539 m/s
3) T = (L+dL)/v = 0.000537505 s
which is not the correct answer.

Thank you for reading and for your feedback,

Josh

 

[PietKuip]

The speed of sound is the square root of the ratio of the elastic modulus to density.
$v=\sqrt{\frac{Y}{\rho}}$

 

[SteamKing]

Homework Statement

From WebAssign:
One mole of nickel (6.02e23 atoms) has a mass of 59 g, and its density is 8.9 g/cm3. You have a bar of nickel 2.44 m long, with a square cross section, 1.8 mm on a side. You hang the rod vertically and attach a 45 kg mass to the bottom, and you observe that the bar becomes 1.6 mm longer. Next you remove the 45 kg mass, place the rod horizontally, and strike one end with a hammer. How much time T will elapse before a microphone at the other end of the bar will detect a disturbance? (Assume a simple cubic lattice for nickel.)

Homework Equations

Young's modulus, diameter of an atom, v=ωd

The Attempt at a Solution

tl;dr - I'm getting 0.000538 seconds, not the right answer.

Atomic Size and Mass:

1) convert given density to kg/m^3 = 8900kg/m^3
2) convert to moles/m^3 (kg/m^3 * mol/kg) = 150847 mol/m^3 (not rounding in my actual calculations)
3) convert to atoms/m^3 (6.022^23 atoms/mol) = 9.084e28 atoms/m^3
4) take the cube root to get the number of atoms per meter, = 4495309334 atoms/m
5) take the reciprocal to get the diameter of an atom, = 2.2245e-10 m/atom
6) find the mass of one atom (kg/mol * mol/atoms) = 9.7974e-26 kg/atom

Young's Modulus: Y=(F/A)/(dL/L)

1) F=mg = (45kg)(9.8N/kg) = 441 N
2) A = (0.0018m)^2 = 3.5344e-6 m^2
3) dL = 0.0016m
4) L = 2.44m
5) Y = 1.834e11 N/m^2

Interatomic Spring Stiffness: Ks,i = dY

1) From above, diameter of one atom = 2.2245e-10 m
2) From above, Y = 1.834e11 N/m^2
3) Ks,i = 40.799 N/m (not rounding in my actual calculations)

Speed of Sound: v = ωd

1) ω = √(Ks,i / m,a)
2) From above, Ks,i = 40.799 N/m
3) From above, m,a = 9.7974e-26 kg
4) ω=2.0406e13 N/m*kg
5) From above, d=2.2245e-10 m
6) v=ωd = 4539 m/s (not rounding in actual calculations)

Time Elapsed:

1) length sound traveled = L+dL = 2.44166 m
2) From above, speed of sound = 4539 m/s
3) T = (L+dL)/v = 0.000537505 s
which is not the correct answer.

Thank you for reading and for your feedback,

Josh

The nickel bar has the 45 kg weight removed from the end and is laid flat before being struck.

Why do you include the change in length due to this weight (which is no longer present) in the calculation of the time it takes for sound to travel thru the rod?

 

[jhyrman]

Thank you PietKuip, that does make it a lot easier. I still get the same result.

SteamKing, I tried it with and without. After rounding to correct sig figs, the result is the same so I know that's not the problem. I wasn't sure how elastic metal is, but I suppose it would go back to it's original length after the weight is removed.

 

[SteamKing]

SteamKing, I tried it with and without. After rounding to correct sig figs, the result is the same so I know that's not the problem. I wasn't sure how elastic metal is, but I suppose it would go back to it's original length after the weight is removed.

Yes, it is reasonable to assume that the bar will return to its original length once the weight is removed.

 

[jhyrman]

Thank you for your help. I found the problem in my math calculating Young's modulus. I corrected that and it finally got the correct answer.