Time it Takes for a Transverse Pulse to Travel a Distance "L" Up a Cable Against Gravity?

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  • #1
domephilis
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Homework Statement
A mass M hangs vertically at the end of a cable of mass m and length L.

(i) How long will it take for a transverse pulse to travel from bottom to top if you ignore m, the cable mass?

(ii) Now repeat, including m and remembering that the velocity of the signal varies with the distance from the bottom end. Show that the answer reduces to part (i) if you take the limit as m goes to 0.

Source: Problem 9.12 of R. Shankar's Fundamentals of Physics I
Relevant Equations
$$v=\sqrt{\frac{T}{\mu}}$$
I had no problem with (i). The tension was ##Mg##. The mass per unit length was ##\frac{m}{L}##. The answer is $$\sqrt{\frac{mL}{Mg}}$$, which was correct.

With (ii), I just tried the same thing but changed the tension to ##(M+m)g##. That did not yield the right answer at all. The solution in the book said that "At any given point along the cable the tension will have to balance both the mass at the end of the cable and any mass from the portion below it (I believe it refers to the travelling pulse)". It then proceeded to set up a differential equation relating the velocity ##\frac{dx}{dt}## to the force at the position of the travelling pulse. This completely negates the premise for the simplified wave equation which assumes that tension is constant across the string for small oscillations. The wave equation my textbook gave is this:$$\frac{\partial^2 \psi}{\partial x^2} = \frac{1}{v^2}\frac{\partial^2 \psi}{\partial t^2}$$. Plus, why did it choose to ignore the force exerted by the ceiling which definitely balances out the weight of the entire mass-string system (not just the bottom part)?
 
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  • #2
domephilis said:
why did it choose to ignore the force exerted by the ceiling
To figure out the velocity at a given point in the string you need to find the tension there. The element at that position does not care how the tension is created.
Write an expression for ##T(x)##.
 
  • #3
haruspex said:
To figure out the velocity at a given point in the string you need to find the tension there. The element at that position does not care how the tension is created.
Write an expression for ##T(x)##.
In deriving the wave equation, the tension was assumed to be constant throughout the string. If we cannot use that “simplified” wave equation to find the wave velocity, I would have to entirely re-derive and solve a different wave equation taking Tension as a function of position this time. Is that what it intends for me to do?
 
  • #4
haruspex said:
To figure out the velocity at a given point in the string you need to find the tension there. The element at that position does not care how the tension is created.
Write an expression for ##T(x)##.
I've since found a resource that explains tension in a bit more detail. I was able to get to the formula in the book by considering the forces acting on each mass element x above the connection with the mass (using the good old free body diagram). I still don't quite understand why we take ##v=\sqrt{\frac{T(x)}{\mu}}## for granted when T is not constant? (Recall that this formula came from wave equation which depended on the simplification that the tension is the same everywhere on the string and only the angle changes).

To elaborate, consider the derivation of the wave equation. I'm going skip the physical reasoning behind the first steps and jump to this: $$\mu \frac{\partial^2 \psi}{\partial t^2} = \frac{\partial T}{\partial x}\frac{\partial \psi}{\partial x}+T(x,t)\frac{\partial^2 \psi}{\partial x^2}$$
where T is the tension on the string as a function of x and t.

Before, we assumed that T is constant in both x and t; ##\frac{\partial T}{\partial x}## is zero; and, we got the canonical formula $$\frac{\partial^2 \psi}{\partial x^2} = \frac{1}{v^2}\frac{\partial^2 \psi}{\partial t^2}$$ That was why $$v = \sqrt{\frac{T}{\mu}}$$

But now, T is not constant. Hence, $$\mu \frac{\partial^2 \psi}{\partial t^2} = \frac{mg}{L}\frac{\partial \psi}{\partial x}+(Mg+x\frac{m}{L}g)\frac{\partial^2 \psi}{\partial x^2}$$. Can we still use d'Alembert's solution to deduce that ##v = \sqrt{\frac{T(x)}{\mu}}##? If so, how?

Thank you for taking the time to help me with this problem. I am quite new to waves and would greatly appreciate the help.
 
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  • #5
domephilis said:
In deriving the wave equation, the tension was assumed to be constant throughout the string.
For the given question, we can obtain from first principles that
$$\frac{\partial^2 \psi}{\partial x^2} = \frac{\mu}{T}\frac{\partial^2 \psi}{\partial t^2}$$
This is by considering F=ma applied to a small element. It makes no assumption about either ##\mu## or ##T## being constant. We could even write:
$$\frac{\partial^2 \psi}{\partial x^2} = \frac{\mu(x,t)}{T(x,t)}\frac{\partial^2 \psi}{\partial t^2}$$
In the special case where they are constant, we get an equation that arises in many contexts and is referred to as "the wave equation". But in this question, ##T=T(x)##.
 
  • #6
haruspex said:
For the given question, we can obtain from first principles that
$$\frac{\partial^2 \psi}{\partial x^2} = \frac{\mu}{T}\frac{\partial^2 \psi}{\partial t^2}$$
This is by considering F=ma applied to a small element. It makes no assumption about either ##\mu## or ##T## being constant. We could even write:
$$\frac{\partial^2 \psi}{\partial x^2} = \frac{\mu(x,t)}{T(x,t)}\frac{\partial^2 \psi}{\partial t^2}$$
In the special case where they are constant, we get an equation that arises in many contexts and is referred to as "the wave equation". But in this question, ##T=T(x)##.
There is an extra ##\frac{\partial T}{\partial x}\frac{\partial \psi}{\partial x}## term. I copied this off of https://personal.math.ubc.ca/~feldman/apps/wave.pdf. This term comes about from the the F part of F=ma. If we ignore the horizontal components of the tension force and only consider the vertical ones (keeping in mind that theta is also a function which depend on x and t), the tension acting on a single section of string is $$T(x+\Delta x,t)\sin(\theta(x+\Delta x,t)) - T(x,t)\sin(\theta(x,t))$$ We can divide this by dx and take the limit as dx goes to 0 and get $$\frac{\partial}{\partial x} [T(x,t)\sin(\theta(x,t))]$$. (Note that this comes from the limit definition of the derivative when applied to the function T multiplied by sin composed with theta) By the product rule, we can get a term which multiplies the first x derivative of T. Since that is zero when T is constant, we can ignore it and achieve the simplified form. But, since it is nonzero in our case, we get the differential equation I've written in the previous post.
 
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  • #7
domephilis said:
There is an extra ##\frac{\partial T}{\partial x}\frac{\partial \psi}{\partial x}## term.
Strictly speaking, yes, but that is eliminated by considering sufficiently small amplitude vibrations. As noted at the page you linked, that gets you to equation (4).
domephilis said:
If we ignore the horizontal components of the tension force and only consider the vertical ones
The string hangs vertically. Equation (4) is obtained by considering transverse forces and acceleration, i.e. horizontal. It is the vertical motion of the element that is ignored. Gravity here only serves to affect the tension and hence, indirectly, the horizontal forces.
The F in equation (4) is an external transverse force. The only external force here is gravity, which is longitudinal. So we end up with what looks like the standard wave equation except that the "c" is a function of x.
 
  • #8
haruspex said:
Strictly speaking, yes, but that is eliminated by considering sufficiently small amplitude vibrations. As noted at the page you linked, that gets you to equation (4).

The string hangs vertically. Equation (4) is obtained by considering transverse forces and acceleration, i.e. horizontal. It is the vertical motion of the element that is ignored. Gravity here only serves to affect the tension and hence, indirectly, the horizontal forces.
The F in equation (4) is an external transverse force. The only external force here is gravity, which is longitudinal. So we end up with what looks like the standard wave equation except that the "c" is a function of x.
I understand that F is an external transverse force which is zero in our case. The text I quoted said that "For small amplitude vibrations, cos(θ) is very close to one and ##\frac{\partial T}{\partial x}## is very close to zero." The latter is clearly not true in our case. Therefore, "the net horizontal force" (which in our case is vertical) is nonzero. Only if it is zero, do we get equation (4). It was valid to ignore the vertical motion of the element if we do not have gravity. My question is : What happens if we take into account this vertical force (which in the case of paper is horizontal), now that we have gravity in our system?
 
  • #9
domephilis said:
Therefore, "the net horizontal force" (which in our case is vertical) is nonzero.
No, it is zero. The small difference in tension between above and below the element balances the weight of the element. The vertical acceleration of the element is negligible for small oscillations.
 
  • #10
haruspex said:
No, it is zero. The small difference in tension between above and below the element balances the weight of the element. The vertical acceleration of the element is negligible for small oscillations.
Since we derived the tension T as a function of x earlier, should we not apply the result ##\frac{\partial T}{\partial x} = \frac{mg}{L}## directly into equation (3)? The reason it went on after equation (3) was that it took T as unknown. But we know T. Unless ##\frac{mg}{L}## is negligibly small (which is not necessarily true), its later argument that the "horizontal" (vertical in our case) is zero doesn't seem to apply in our circumstances since $$\frac{\partial}{\partial x}[T(x,t)\cos(\theta(x,t))] = \frac{\partial T}{\partial x} \cos(\theta(x,t)) - T(x,t)\frac{\partial \theta}{\partial x}\sin(\theta(x,t)) \approx \frac{mg}{L}$$. I do apologize if I seem a bit slow here. I am perfectly willing to concede that we can ignore certain negligible effects, but I would just like to know why this is negligible.
 
  • #11
domephilis said:
In deriving the wave equation, the tension was assumed to be constant throughout the string. If we cannot use that “simplified” wave equation to find the wave velocity, I would have to entirely re-derive and solve a different wave equation taking Tension as a function of position this time. Is that what it intends for me to do?
Since Shankar’s textbook is for a first course in physics at the college level, I’m confident the student is not expected to justify ##v = \sqrt{T(x)/\mu}## when the tension varies with ##x##. I’ve seen this problem in several textbooks at the same level and none of them give any derivation or justification. I think that you are expected to just "naturally assume" that ##v = \sqrt{T/\mu}## for uniform tension carries over to ##v = \sqrt{T(x)/\mu}## when ##T## is a function of ##x##. The fact that you are uncomfortable with this is a good sign!

I did some searching and found a treatment in the old, classic text A Treatise on the Dynamics of a System of Rigid Bodies by Edward Routh. This book went through at least 6 editions in the late 1800’s and early 1900’s. This is an advanced book! It treats the hanging cable and derives ##v = \sqrt{T(x)/\mu}##. If you want to take a look, here is a link. The analysis is in section 536 which starts at the bottom of page 307. The wave equation is given in equation (4) and agrees with the wave equation you cited in post #4. (Routh uses ##m## for the mass density.)

Then Routh gives a clever derivation of the wave speed. A wave pulse will not keep its shape since different parts of the pulse travel at different speeds. But he shows that if ##a## and ##b## mark the upper and lower extremities of the pulse, then the wave speed of ##a## is given by ##v_a = \sqrt{T(x_a)/\mu}## and the speed of ##b## is ##v_b = \sqrt{T(x_b)/\mu}##, where ##x_a## and ##x_b## are the instantaneous locations of ##a## and ##b##. For a short pulse in which ##a## and ##b## are close together, ##v_a \approx v_b##.
 
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  • #12
TSny said:
Since Shankar’s textbook is for a first course in physics at the college level, I’m confident the student is not expected to justify ##v = \sqrt{T(x)/\mu}## when the tension varies with ##x##. I’ve seen this problem in several textbooks at the same level and none of them give any derivation or justification. I think that you are expected to just "naturally assume" that ##v = \sqrt{T/\mu}## for uniform tension carries over to ##v = \sqrt{T(x)/\mu}## when ##T## is a function of ##x##. The fact that you are uncomfortable with this is a good sign!

I did some searching and found a treatment in the old, classic text Dynamics of Rigid Bodies by Edward Routh. This book went through at least 6 editions in the late 1800’s and early 1900’s. This is an advanced book! It treats the hanging cable and derives ##v = \sqrt{T(x)/\mu}##. If you want to take a look, here is a link. The analysis is in section 536 which starts at the bottom of page 307. The wave equation is given in equation (4) and agrees with the wave equation you cited in post #4. (Routh uses ##m## for the mass density.)

Then Routh gives a clever derivation of the wave speed. A wave pulse will not keep its shape since different parts of the pulse travel at different speeds. But he shows that if ##a## and ##b## mark the upper and lower extremities of the pulse, then the wave speed of ##a## is given by ##v_a = \sqrt{T(x_a)/\mu}## and the speed of ##b## is ##v_b = \sqrt{T(x_b)/\mu}##, where ##x_a## and ##x_b## are the instantaneous locations of ##a## and ##b##. For a short pulse in which ##a## and ##b## are close together, ##v_a \approx v_b##.
Wonderful. I think this exactly what I need to satisfy my little curiosity. Thank you very much for this. I will definitely take some time to digest this.
 
  • #13
domephilis said:
Since we derived the tension T as a function of x earlier, should we not apply the result ##\frac{\partial T}{\partial x} = \frac{mg}{L}## directly into equation (3)? The reason it went on after equation (3) was that it took T as unknown. But we know T. Unless ##\frac{mg}{L}## is negligibly small (which is not necessarily true), its later argument that the "horizontal" (vertical in our case) is zero doesn't seem to apply in our circumstances since $$\frac{\partial}{\partial x}[T(x,t)\cos(\theta(x,t))] = \frac{\partial T}{\partial x} \cos(\theta(x,t)) - T(x,t)\frac{\partial \theta}{\partial x}\sin(\theta(x,t)) \approx \frac{mg}{L}$$. I do apologize if I seem a bit slow here. I am perfectly willing to concede that we can ignore certain negligible effects, but I would just like to know why this is negligible.
The linked article says:
Then the net horizontal force on it must be zero. That is,
T (x + ∆x, t) cos θ(x + ∆x, t) − T (x, t) cos θ(x, t) = 0
But the case considered there has no external force in that direction. For the case in this thread, gravity acts in that direction, so the equation becomes (taking x as increasing upwards)

T (x + ∆x, t) cos θ(x + ∆x, t) − T (x, t) cos θ(x, t) = μΔxg
leading to ##\frac{\partial T}{\partial x}=\mu g## for small θ.
 
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  • #14
haruspex said:
The linked article says:
Then the net horizontal force on it must be zero. That is,
T (x + ∆x, t) cos θ(x + ∆x, t) − T (x, t) cos θ(x, t) = 0
But the case considered there has no external force in that direction. For the case in this thread, gravity acts in that direction, so the equation becomes (taking x as increasing upwards)

T (x + ∆x, t) cos θ(x + ∆x, t) − T (x, t) cos θ(x, t) = μΔxg
leading to ##\frac{\partial T}{\partial x}=\mu g## for small θ.
Yes, I agree. I think that’s why the wave equation assumes a different form. As TSny points out, we can still deduce the wave velocity by only considering the boundary points of the pulse which has the special properties that the horizontal velocity (time derivative) is zero and the angle is 0 relative to vertical (hence ##\frac{d \psi}{dx}=0##). The resource TSny quoted (by Edward Routh) derives this in greater detail, but the end result ##v=\frac{T(x)}{\mu}## at the special points resolves my confusion about the solution offered by the book.
 

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