# Longitudinal waves

1. Sep 8, 2014

### Karol

1. The problem statement, all variables and given/known data
A steel pipe of length 60m is hit at one side. a man standing at the other side hears 2 sounds, one that passes in the pipe and one through the air. what is the time difference.
Take the modulus of elasticity as 2,180,000[kgf/cm2]

2. Relevant equations
Velocity in the pipe: $u=\sqrt{\frac{E}{\rho}}$

3. The attempt at a solution
I assume the specific mass: 8[gr/cm3]=0.008[kg/cm3]=0.0008[hyl/cm3]
$$u=\sqrt{\frac{2,180,000}{0.0008}}=52,202[cm/sec]=522[m/sec]$$
$$\delta v=522-346=176$$
$$60=176\cdot t\rightarrow t=0.34[sec]$$
It should be 0.16[sec]

2. Sep 8, 2014

### Staff: Mentor

You calculated the time it takes to get a distance of 60m between the two sound waves, not the time difference at 60m distance.

How long does the sound in steel need to reach the other end? And the sound in air?

That's a weird type of steel, typical sound velocities are a factor of 10 larger.

3. Sep 9, 2014

### Karol

$$\frac{60}{346}-\frac{60}{522}=0.06$$
And it's not good

4. Sep 9, 2014

### BvU

No, but it's a lot better!

Could you work in decent units to get m/s as speed ? That might help quite a bit and do justice to mfb's hint.
(would also help me, because I collect exotic units and can't find the hyl anywhere)

5. Sep 9, 2014

### Karol

I'm sure it's a problem of units because if i take ρ 10 times smaller, ρ=8E-5, the final answer is almost perfect, 0.14. and it also gives a much higher speed in the steel.
If i take values from the literature:
E=20E10[N/m2]
ρ=8000[kg/m3]
$$u=\sqrt{\frac{20E10}{8000}}=5000[m/s]$$
Gives the exact answer. but i don't know where is my mistake.
The hyl is the mass unit in the technical or gravitational system of units, the one with the kgf. it's 9.8 kilograms.
I was given the value of 2,180,000[kgf/cm2] and it's right also according to the literature, so, the problem must be with the ρ. i have to take it 10 times smaller than [kg/cm3] if i want to use those technical units and i did take it 10 times smaller:
8[gr/cm3]=0.008[kg/cm3]
And i took 0.0008, but i should have taken 0.00008, why?

6. Sep 9, 2014

### BvU

Well, thanks for the hyl. Funny they should express it in kilograms, but I suppose outside of the realm of physics anything goes. Within physics, even the kgf [STRIKE]should be [/STRIKE] is forbidden.

If you want to persist in this, you still need the factor g (9.80665 m/s2). It's the law (or rather, the relationship c = √(Y/ρ)). You create your own disaster by dividing kgf by kg.

There shall be no fudging with the density of steel.

The SI unit of Y is Pa, the dimension is N/m2 or kg/(m s2). Divide by kg/m3) to get m2/s2 which is good indeed. Usually Y is expressed in GPa by those born after the 19th century.

7. Sep 10, 2014

### Karol

Yes but i still don't understand where i am wrong. i want to use my (technical) dimensions, and indeed i divided ρ once by 10, why should i divide it by 100?

8. Sep 10, 2014

### BvU

You can actually use your technical dimensions, but then you need a different expression for the speed of sound. It's like if you use stone/(acrefoot) for $\rho$ and still want the speed in furlongs/fortnight.

9. Sep 10, 2014

### Karol

I am beginning to understand, but why do i have to use a different expression? indeed i use consistent dimensions. and in both, the SI and the technical system the speed is in [m/s]?

10. Sep 10, 2014

### BvU

Your hyl has a g in it, your $\rho$ has not. So somehow this has to be accounted for.

Be assured: you're not the only one with this kind of problem: people who use lbf for force have the same handicap. I wonder if they are also forced to use a different expression. Anyone from this ancient tribe that has to live with non-SI systems care to comment ?

And don't bother to propose other systems than SI. I know setting c=1 and h=1 (or was it $\hbar$?) is much better, but it just isn't practical in the grocery store.

11. Sep 10, 2014

### Staff: Mentor

Measuring weight in lb wt meant measuring mass in slugs, IIRC. The same equation holds,

W = m g

12. Sep 10, 2014

### Karol

Even with nearly SI units i don't succeed:
E=20E10[N/m2]=20E6[N/cm2]
$\rho$=8[gr/cm3]=0.008[kg/cm3]
$$u=\sqrt{\frac{20E6}{0.008}}=50,000[cm/s]=500[m/s]$$
And it's 10 times smaller than reality

13. Sep 10, 2014

### olivermsun

1 N = 1 kg m s$^{-2}$

The extra 10$^2$ in the cm-to-m conversion will give you a factor of 10 when you take the square root in your velocity equation:

$$u = \sqrt{\dfrac{E}{\rho}} = \sqrt{ \dfrac{200 \mathrm{\,GPa}}{8 \times 10^3 \mathrm{\,kg\,m}^{-3}} } = \sqrt{ \dfrac{200 \times 10^{9} \mathrm{\,N\,m}^{-2}}{8 \times 10^3 \mathrm{\,kg\,m}^{-3}} } = \sqrt{25 \times 10^{6} \,\mathrm{ \dfrac{\left(kg\,m\,s^{-2}\right)m}{\,kg} } } = 5 \times 10^{3} \,\mathrm{m/s} .$$

Last edited: Sep 10, 2014
14. Sep 10, 2014

### Karol

This i wrote in my post, i used SI units and got u=5000[m/s]
I want to use, for the sake of understanding and not because i have to or am used to, other units.
It seems, if i understand right, that for every unit system other than SI i have to use a different equation, is it right? it doesn't sound logical, and the equation $u=\sqrt{\frac{E}{\rho}}$ isn't developed with connection to any system of units.

15. Sep 10, 2014

### nasu

But you need to use consistent units to get valid results.
You are using a mixture of CGS and SI.
The SI unit for force contains a meter in it as shown by olivermsun already.
So does not go with the cm^3 in the density.

If you want CGS units use the CGS unit for force (it's dyne). If you want to use Newtons, use the SI unit for density. This is all it is to it.
The value of the Young's modulus in CGS will be 2x10^12 dyn/cm^2 and the density will be 8 g/cm^3. The results will be in cm/s.

16. Sep 10, 2014

### Karol

Why do i have to use, as you say, consistent units? just because they are called, for example, SI?
Isn't it good that i use, for example, centimeters instead of meters all over the equation, in all places? indeed i don't mix between centimeters and meters, for example. i don't use meters in the numerator and centimeters in the denominator, i use centimeters in both.
But i think the question is harder about other units, the mass and force, since in the numerator i use newtons and in the denominator kg, like in SI, but who said it must be? maybe i have to use kgf and kg in order that it come out like in SI?

17. Sep 10, 2014

### nasu

But you DID mix units, as was already mentioned a couple of times.
You did not use cm all over the equation.
Using N and cm in the same formula you are mixing m and cm.
Newton is just a shortcut for kg*m/s^2. The meter is there even if you use the "newton" to wrap it in.

18. Sep 10, 2014

### Staff: Mentor

No, the SI system is just the easiest system to use. All unit systems allow to be consistent.
The problem is as easy as $\frac{m}{cm} \neq 1$. Your calculation assumes you could cancel those in the fractions, but you cannot - unless you convert it first: $\frac{m}{cm} = \frac{100cm}{cm}=100$. Sometimes the units are "hidden", like the meter in the Newton definition, or m/s^2 from g in the kg/kgf conversion.

If you keep in mind that kgf and kg are different units, and convert if necessary, it is fine.

19. Sep 10, 2014

### Karol

I thank you, nasu, very much for this answer, now i understand.
And it also solved the issue with the [kgf/cm2] and [hyl/cm3] units, because hyl is also defined with m, thanks!