Understanding Waves on a String

  • #1

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Hello I am having trouble understanding why the horizontal component Fx is equal to the F for a wave on a string I am trying to understand Energy in wave motion. The picture attached is a representation of my problem. My textbook tells me that this is because it is a transverse wave which means the the direction of the wave is perpendicular to the displacement of the medium. I still don't understand why this is.

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  • #2
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It is just a printing error. F on the left side should say F1x and F on the right side should be F2x.
 
  • #3
Delta2
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Since the wave is transverse, the motion of each segment is only on the vertical y axis, hence on the horizontal x axis there is no motion, so according to newton's second law for the x-axis we ll have
##\sum F_x=ma_x=0## (that is, the total force on x-axis must be zero since there is no movement there, that is ##a_x=0##)

therefore ##F_{1x}-F_{2x}=0\Rightarrow F_{1x}=F_{2x}##

Your book seems to "bypass" this explanation and just sets ##F_{1x}=F_{2x}=F##
 
  • #4
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Regarding your other questions, waves can be confusing, so don’t feel like it should be obvious. There is a lot going on.

It’s hard to tell from your question exactly what you would like explained. If the primary problem is understanding what they mean by “transverse” the easiest way to grasp that is by comparing to the other option “longitudinal”. And the quickest way to grasp that is by seeing it. Buy a cheap plastic slinky with a reasonably large diameter. Better yet buy 3 and tape together the last coils to make one long slinky. Get a friend to hold one end and stretch it out on the floor. Make a quick side to side motion. A transverse pulse will travel down the slinky. Now make a quick forward and back movement. A longitudinal pulse will travel down the slinky. You will see what they mean about the motion of the particles being perpendicular or parallel to the direction of propagation.
 
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haruspex
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It is just a printing error. F on the left side should say F1x and F on the right side should be F2x.
No, it is intentional that the label is the same. The accompanying text explains this. It might have been better to label both Fx.
 
  • #6
Orodruin
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Since the wave is transverse, the motion of each segment is only on the vertical y axis, hence on the horizontal x axis there is no motion, so according to newton's second law for the x-axis we ll have
##\sum F_x=ma_x=0## (that is, the total force on x-axis must be zero since there is no movement there, that is ##a_x=0##)

therefore ##F_{1x}-F_{2x}=0\Rightarrow F_{1x}=F_{2x}##

Your book seems to "bypass" this explanation and just sets ##F_{1x}=F_{2x}=F##
This is actually not as obvious as it may seem at first. The wave being purely transversal is often introduced by assumption and the x-components cancelling out is then a result of an assumption. However, you can show more rigorously that this is approximately true for small oscillations. If you have large oscillations you need to consider interactions between longitudinal and transversal modes, which are non-linear and therefore do not appear at leading order for small oscillations.

Edit: To be more precise. Unless I did an arithmetic error somewhere, the differential equation for the transverse displacement with the leading non-linear terms is
$$
\rho u_{tt} - T_0 u_{xx} = - \partial_x [(\lambda-T_0)\epsilon u_x] + \mathcal O(\epsilon^2 u_x)
$$
where
$$
\epsilon = \sqrt{(1+v_x)^2 + u_x^2} - 1
$$
is the strain in the string, ##v## is the longitudinal displacement, ##T_0## the tension at equilibrium and ##\lambda## the constant from Hooke's law. When the right-hand side is negligible, we recover the wave equation.
 
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  • #7
sophiecentaur
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Regarding your other questions, waves can be confusing, so don’t feel like it should be obvious. There is a lot going on.

It’s hard to tell from your question exactly what you would like explained. If the primary problem is understanding what they mean by “transverse” the easiest way to grasp that is by comparing to the other option “longitudinal”. And the quickest way to grasp that is by seeing it. Buy a cheap plastic slinky with a reasonably large diameter. Better yet buy 3 and tape together the last coils to make one long slinky. Get a friend to hold one end and stretch it out on the floor. Make a quick side to side motion. A transverse pulse will travel down the slinky. Now make a quick forward and back movement. A longitudinal pulse will travel down the slinky. You will see what they mean about the motion of the particles being perpendicular or parallel to the direction of propagation.
Yes. This is confusing and it's due to the attempt to 'classify at all costs'. Fact is that, whenever there is a transverse displacement, the string has to stretch a bit somewhere and the tension will not actually be the same and there will be some longitudinal component. The speed of longitudinal wave propagation is much higher than the transverse wave speed so the 'stretch' is only slight and it closely follows the transverse displacements. With a slinky, the stretch is significant because the longitudinal speed is quite comparable (You can see the bunching and stretching of the coils on a transverse slinky wave. Surface water waves are the same and peaks / troughs on the surface have to involve motion of the water back and forth. Surface waves are very much not only transverse, despite the fact that are the prime example that's always given.
 
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Yes. This is confusing and it's due to the attempt to 'classify at all costs'. Fact is that, whenever there is a transverse displacement, the string has to stretch a bit somewhere and the tension will not actually be the same and there will be some longitudinal component. The speed of longitudinal wave propagation is much higher than the transverse wave speed so the 'stretch' is only slight and it closely follows the transverse displacements. With a slinky, the stretch is significant because the longitudinal speed is quite comparable (You can see the bunching and stretching of the coils on a transverse slinky wave. Surface water waves are the same and peaks / troughs on the surface have to involve motion of the water back and forth. Surface waves are very much not only transverse, despite the fact that are the prime example that's always given.
Actually, let me expand a bit on that. In the case of the string, the transverse wave velocity is given by ##\sqrt{T_0/\rho}## (as seen in my previous post). On the other hand, if you do the corresponding math for the longitudinal wave, you would find that the linearised equation is given by
$$
\rho v_{tt} - \lambda v_{xx} = 0
$$
so the longitudinal wave velocity is instead ##\sqrt{\lambda/\rho}##. The rest tension ##T_0## and ##\lambda## are related through Hooke's law ##T_0 = \lambda \epsilon_0##, where ##\epsilon_0## is the strain in the string at equilibrium, which typically satisfies ##\epsilon_0 \ll 1## and therefore ##\lambda \gg T_0##.
 
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  • #9
Delta2
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Yes, fine, strictly speaking there is a longitudinal wave too, so there is motion in the x-axis involved as well, but for small oscillations the longitudinal wave is really negligible so the balance of forces in the x-axis holds as a very good approximation.
 
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  • #10
Since the wave is transverse, the motion of each segment is only on the vertical y axis, hence on the horizontal x axis there is no motion, so according to newton's second law for the x-axis we ll have
##\sum F_x=ma_x=0## (that is, the total force on x-axis must be zero since there is no movement there, that is ##a_x=0##)

therefore ##F_{1x}-F_{2x}=0\Rightarrow F_{1x}=F_{2x}##

Your book seems to "bypass" this explanation and just sets ##F_{1x}=F_{2x}=F##
Thank you all for your help. This has clarified it bit for me. However I do not understand why the x component is equal to the tension F and why the y component doesn't play a role in that? from what I remember the square root of the x and y components squared gives the magnitude.
 
  • #11
Delta2
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Thank you all for your help. This has clarified it bit for me. However I do not understand why the x component is equal to the tension F and why the y component doesn't play a role in that? from what I remember the square root of the x and y components squared gives the magnitude.
By F here we mean just the horizontal component of the tension of the string (the full tension is ##\sqrt{F^2+{F_y}^2}## ), which we take to be (by approximation) constant throughout the string length. This approximation is valid only for small oscillations. For large oscillations , F varies with the string length x and therefore there is motion also in the x-axis, that is a longitudinal wave.
 
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