Given linear density of two strings of total length 4m, find total length.

[chris_avfc]

Homework Statement

If you were given the
- linear density 1 = 2g/m
- linear density 2 = 4g/,
- total length of the strings = 4m
- Length 1 > Length 2
- the fact that when a pulse is sent from a knot in the strings it reaches the ends at the same time

How would you go about working out the length of each string?

The Attempt at a Solution

I've been trying to use
μ = m/l
v = (T/μ)^0.5
v = s/t

I tried equating the times as they should be the same, but I get stuck after some substitution.

Any ideas?

Thanks

 

[grzz]

'...when a pulse is sent from a knot in the strings it reaches the ends at the same time ...'

You mean the knot tying the two strings together.Ok?

 

[chris_avfc]

'...when a pulse is sent from a knot in the strings it reaches the ends at the same time ...'

You mean the knot tying the two strings together.Ok?

Yeah that's it, the two strings are tied together, didn't make that very clear.

I got to a point where I have
L_1 (T_2/μ_2)^0.5 = L_2 (T_1/μ_1)^0.5

(_ = Subscript)

So I have a feeling I've gone wrong somewhere.

 

[grzz]

..L_1 (T_2/μ_2)^0.5 = L_2 (T_1/μ_1)^0.5...

You have L x speed. Does that give 'time'?

 

[chris_avfc]

You have L x speed. Does that give 'time'?

Yeah, but I only have the μ values to substitute into there.
I've been looking through the textbook at all the equations, but I'm struggling to get anywhere with it.

 

[grzz]

Yeah, but I only have the μ values to substitute into there.
...

You have also to substitute the different lengths of the strings.

 

[chris_avfc]

You have also to substitute the different lengths of the strings.

I've substituted L for (m/μ)

I tried substituting it in both sides, but that of course eliminates both L's so that would be no use.
I tried substituting in for one side and I end up with L_2 = ((( T_2)(m_1^2))/T_1)^0.5
( I realize that isn't very clear but its a fraction which is all square rooted. )

I'm going to have to look for some Tension and Mass equations then, as neither values are given.

Cheer btw.

 

[grzz]

...I tried substituting it in both sides, but that of course eliminates both L's so that would be no use.
I tried substituting in for one side and I end up with L_2 = ((( T_2)(m_1^2))/T_1)^0.5
...

Try to THINK on what you are doing. Do not substitue by 'trial and error'.
The length of string is 4m. So try substituting x for one side and (4-x) for the other side.

 

[chris_avfc]

Try to THINK on what you are doing. Do not substitue by 'trial and error'.
The length of string is 4m. So try substituting x for one side and (4-x) for the other side.

Yeah that would make a lot more sense, I should have thought have that.

But I still have a problem with the unknown tensions?

 

[grzz]

What happens if the tension in one string is greater than the tension in the other?

 

[chris_avfc]

What happens if the tension in one string is greater than the tension in the other?

In this case, if its the tension of the second string is more than twice the tension in the first string, the wave will travel faster down it?

 

[grzz]

I guess that if the tension is unequal at some point then the system will ... break up at that point. What I mean is that the tension is the same throughout.

 

[chris_avfc]

I guess that if the tension is unequal at some point then the system will ... break up at that point. What I mean is that the tension is the same throughout.

okay, if I'm doing this right I'm left with x = (-32/T)^0.5

 

[grzz]

okay, if I'm doing this right I'm left with x = (-32/T)^0.5

That cannot be right because T ought to cancel out and x cannot involve the square root of a negative number.

If one shows the working in deriving the equation for x one can give better help.

 

[chris_avfc]

That cannot be right because T ought to cancel out and x cannot involve the square root of a negative number.

If one shows the working in deriving the equation for x one can give better help.

yeah that's what I though as well, so I'm sure its wrong somewhere.

Anyway here is my working out at the moment

 

[grzz]

AS SOON as the tensions T1 and T2 appear, on your first page, just cancell them out because T1=T2.

 

[chris_avfc]

AS SOON as the tensions T1 and T2 appear, on your first page, just cancell them out because T1=T2.

So I should make it (1/μ_1)^0.5 and (1/μ_2)^0.5?

 

[grzz]

v=dist/time

so time = dist/vel

time 1 = time 2

L1/v1 = l2/v2
L1 *v2 = L2*v1

Now substitute for v1 and for v2, THEN cancell T1 and T2.

 

[chris_avfc]

v=dist/time

so time = dist/vel

time 1 = time 2

L1/v1 = l2/v2
L1 *v2 = L2*v1

Now substitute for v1 and for v2, THEN cancell T1 and T2.

L1 *v2 = L2 *v1

L1 * (T2/μ2)^0.5 = L2 * (T1/μ1)^0.5

so when you say cancel the T do you mean?

L1 * (T2)^0.5 * (1/μ2)^0.5 = L2 * (T1)^0.5 * (1/μ1)^0.5
L1 * (T)^0.5 * (1/μ2)^0.5 = L2 * (T)^0.5 * (1/μ1)^0.5
L1 * (1/μ2)^0.5 = L2 * (1/μ1)^0.5