wanted to ask as mentioned above that how can energy transfer from the first node established in a standing wave by the superposition of two equal and oppositely directed waves,that is when the waves come in contact for the first time to form a node how does the second wave travel past the first one and establish other nodes,because the nodes according to my text book have zero displacement thus zero kinetic energy and zero movement
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7$\begingroup$ Question for the student: what is the net energy transfer of a standing wave? $\endgroup$– dmckee --- ex-moderator kittenCommented Oct 5, 2015 at 15:18
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6 Answers
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There are two effects to bear in mind here. Firstly, the standing wave does not simply spring into existence wholly formed- it evolves from a travelling wave. The second is that the zero amplitude of the vibration at the node does not imply that energy cannot be transferred beyond it. In a transverse wave the node acts as a pivot, rather like the pivot of a see-saw- the pivot doesn't move but nonetheless a force applied at one end of the seesaw causes movement at the other end beyond the stationary pivot.
To understand the role of a node in a longitudinal wave, consider an instance of Newton's cradle with three balls- if one of the end balls is raised and allowed to fall its momentum is transferred to the ball at the other end even though the central ball does not move with a noticeable amplitude. Here the central ball is acting like the node- a stationary element through which momentum and energy can be transferred.
answered Sep 27, 2019 at 14:20
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$\begingroup$ In the case of a see-saw, energy is transferred at the pivot via the flexure of the board. In Newton's cradle, energy is transferred via elastic deformation of the center ball. $\endgroup$ Commented Sep 29, 2019 at 15:58
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$\begingroup$ Correct. They are examples of two mechanisms that allow periodic motions with a large amplitude to be propagated through a point with negligible amplitude of movement. The OP's conceptual difficulty seemed to arise from the assumption that such propagation wasn't possible. $\endgroup$ Commented Sep 29, 2019 at 17:41
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$\begingroup$ It should be noted though that when there is precisely equal energy flow in both directions, there is no net energy flow at the nodes. $\endgroup$ Commented Sep 29, 2019 at 20:45
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the standing wave is just an observed outcome of two waves of the same frequency traveling in opposite directions and interfering with one another. The energy in the two waves isn't traveling 'up and down' but rather in the direction of the two waves. When the two waves meet and are out of phase by 180 degrees they 'cancel' one another and appear as a node. When they are in phase with one another they add and the peak moves up and down with the frequency of the component waves.
answered Oct 5, 2015 at 15:28
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$\begingroup$ but when the on the first interference the waves w1 and w2 cancel out each other and form a node, than how will wave w2 and energy past that node and cancel out the the wave w1 at other to form other nodes??? $\endgroup$– BatwayneCommented Oct 5, 2015 at 15:51
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1$\begingroup$ The net amplitude cancels, not the energy. $\endgroup$ Commented Oct 5, 2015 at 17:32
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$\begingroup$ but energy from one particle is transmitted to another particle by motion that is amplitude so how come enrgy will be transfered to the next particle $\endgroup$– BatwayneCommented Oct 6, 2015 at 10:29
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A standing wave is a "stationary" or a "steady-state" picture established after many wavelengths have interfered.
Bofore that, there is a non stationary motion that carries energy.
Note, the energy density is calculated from the summary amplitude, not as a sum of two energies of the opposite waves :-)
answered Aug 15, 2019 at 14:18
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$\begingroup$ Yes, that's the energy $density$. But the $rate\ of\ energy\ propagation$ can be calculated as the algebraic sum (taking account of direction of flow) of energy propagation rates of the two progressive waves. This sum is zero all along the path of the waves. $\endgroup$ Commented Sep 27, 2019 at 13:46
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$\begingroup$ @PhilipWood: Define, please, the rate of energy propagation. Is it a vector? Is it a gradient of energy density? $\endgroup$ Commented Sep 27, 2019 at 16:43
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$\begingroup$ It's the surface integral $\int_Su\ \vec{v}.\vec{dS}$ for a wave of energy density 𝑢 propagating with velocity 𝑣⃗ through an area $S$. So, like magnetic flux, it is not a vector, but can be positive or negative (dependent upon which of the two normals we choose for $\vec{dS}$. But please accept an apology. I'm now convinced that energy fluxes don't sum in the manner I suggested. $\endgroup$ Commented Sep 28, 2019 at 11:52
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Typically in a wave, energy exists in two forms: potential and kinetic. In the case of a stretched string, potential energy at any point on the string corresponds to displacement from the straight line connecting the two ends, while kinetic energy corresponds to the speed at which the string is moving at each point along the string.
At a node in a "perfect" standing wave on a vibrating string, the string is neither displaced nor moving, so there is no energy at that point. No energy passes through that point either. Of course, if each of two counter-propagating waves is considered separately, energy is propagating in each. But in a perfect standing wave the energy propagation of the two waves is equal and opposite.
However, there is never a "perfect" standing wave. Because there are always energy losses due to friction, etc., the counter-propagating waves do not perfectly cancel each other at the nodes (which in this case are points of energy minima, not points of zero energy). If the resulting standing wave is analyzed in detail, it can be seen that energy does flow through those minimal but nonzero nodes.
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The nodes are zeros in the resultant (standing) wave. They have no effect on the propagation of the constituent progressive waves. They arise by antiphase superposition of the progressive waves, whose (scalar) amplitude is non-zero and the same all along the path of the waves, including at the nodes themselves! [That's the key thing about the Principle of Superposition: a wave propagates just as if other waves weren't there!]
Of course we can't observe the progressive waves, only their resultant, the standing wave. Indeed, we don't always even have to think of a standing wave as the superposition of progressive waves travelling in opposite directions; we can regard it simply as a collection of oscillations whose amplitude varies smoothly from point to point, with zeros at nodes, and with phase reversals at each node.
answered Sep 27, 2019 at 13:14
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Standing waves are poorly taught and you are correct to wonder how the energy travels past the first node. The classical example I give to my students is: imagine two speakers facing each other. When you turn them on, waves travels toward each other. Do you expect the waves to collide and stay at the collision point or do you expect the waves to keep traveling? Energy is not travelling in the standing wave they told you. So, what happens to the energy that is continuously poured into space by the two speakers? Where does it go? Now turn off the speakers, since you have created a "standing wave" nothing should happened, what was there stays there, is it what you have observed?
Obviously all the above does not happen and the simple explanation of standing waves you can read in elementary courses is just bogus.
Nothing is standing in standing waves. At all time the two waves keep traveling from right to left and from left to right, and the energy from wave 1 and 2 keeps also traveling in the two directions. However, the two waves carry the same amount of energy, so if you consider a volume between the two sources, its net time-average gain or loss of energy is null.
