Find Energy of Antinode & Node in Standing Wave of Time Period T

  • Thread starter Thread starter varunika
  • Start date Start date
  • Tags Tags
    Nodes
AI Thread Summary
In a standing wave, nodes have zero amplitude and antinodes have maximum amplitude, leading to different energy distributions. The kinetic energy at a node is zero, while the potential energy at an antinode is also zero, indicating that energy oscillates between kinetic and potential forms. To find when the energy at a node equals that at an antinode, one must equate the kinetic energy of the node to the potential energy of the antinode. The wave speed can be expressed in terms of tension and mass per unit length, and the wave function can be represented using sinusoidal functions. Understanding these relationships is crucial for solving the problem of energy equivalence in standing waves.
varunika
Messages
19
Reaction score
0
a standing wave of time period T is set up in a string clamped between two rigid supports. at t=0 antinode is at its maximum diaplacement 2A. at what time will the energy of antinode be equal to that of a node?


what i know is...
that node has zero amplitude..n antinode has maximum amplitude. and that the maximum potential energy= 1/2*ρ*ω2A2. which also equals the kinetic energy..
moreover it can be seen that the amplitude of the given wave is 2A.
T=2∏/ω.
but i am unable to relate the above information in order to solve the given query.
please tell me how can i find the energy of an antinode and a node?

i am very weak in physics... please guide me through it.
 
Physics news on Phys.org
I assume by 'energy' of a point on the string the question means the sum of KE and PE at that point.
What form does the PE take in a vibrating string? How will that be distributed along the string?
 
energy of a point on a string= 1/2Δmvy2 + 1/2Δmω2y2 (where y is the amplitude of that point)
so this means at a node total energy woul become equal to 1/2Δmvy2 since y=0 and the total energy at the anti node would become equal to 1/2Δmvy2 + 1/2Δmω2(2A)2.

am i right about this?

and if so... how do i get to know the velocity at the node and antinode? and how do i relate this to the timepriod of the wave?

is the velocity of an anti node zero?

moreover value of ω is also not given..

please please guide me more..
 
The velocity at the nodes is zero since they don't move. The formula you provided for the energy looks right but it isn't.
 
dauto said:
The velocity at the nodes is zero since they don't move. The formula you provided for the energy looks right but it isn't.

then how do i find the formula? i searched it all over.. and even thought as much as i could... but i am getting nowhere :cry:
 
Your expression for the kinetic energy is correct but the expression for the potential energy isn't. The potential energy is given by the stretching of the spring, not by its displacement.

it is proportional to the square of the derivative of the wave function (dy/dx)2
 
dauto said:
Your expression for the kinetic energy is correct but the expression for the potential energy isn't. The potential energy is given by the stretching of the spring, not by its displacement.

it is proportional to the square of the derivative of the wave function (dy/dx)2

sir... but the following link says exactly what i said: http://hyperphysics.phy-astr.gsu.edu/hbase/waves/powstr.html (please refer to the section: "energy in a string wave" of this link)
 
varunika said:
sir... but the following link says exactly what i said: http://hyperphysics.phy-astr.gsu.edu/hbase/waves/powstr.html (please refer to the section: "energy in a string wave" of this link)

That page is wrong, unfortunately. The final result for the total energy propagation is correct (for a traveling harmonic wave), but the details of the derivation are not.
 
so according to you.. the potential energy should be equal to 1/2Δmω2(dy/dx)2 ??
 
  • #10
dauto said:
The velocity at the nodes is zero since they don't move.

this means the total energy at the node is zero.. am i right?
 
  • #11
varunika said:
so according to you.. the potential energy should be equal to 1/2Δmω2(dy/dx)2 ??

Almost, but not quite. It's given by

ΔW = (1/2) Δm v2 (dy/dx)2

where v is the wave speed.
 
  • #12
dauto said:
The velocity at the nodes is zero since they don't move.
this means energy at a node is zero. am i right?
 
  • #13
varunika said:
this means energy at a node is zero. am i right?

No, the potential energy at the anti-node is always zero and the kinetic energy at the node is always zero. The energy flips back and forth between Kinetic and Potential as it bounces back and forth between the anti-node and the node.
 
  • #14
so... back to the asked question..
in the query.. we have to find the time wen energy at node=energy at anti node.
KE of node = PE of anti node
∴ i/2Δmv2 = 1/2ΔmV(dy/dx)2
v=√T/μ
would i be wrong if i relate this in the equation... since we must need to relate the timeperiod with energy?
 
  • #15
dauto said:
No, the potential energy at the anti-node is always zero and the kinetic energy at the node is always zero. The energy flips back and forth between Kinetic and Potential as it bounces back and forth between the anti-node and the node.

so... back to the asked question..
in the query.. we have to find the time wen energy at node=energy at anti node.
KE of node = PE of anti node
∴ 1/2Δmv2 = 1/2ΔmV(dy/dx)2
v=λf and f=1/T
∴ v=λ/T
would i be wrong if i relate this in the equation... since we must need to relate the timeperiod with energy?
and my other doubt is... that equation of wave is not given... so how to find dy/dx and λ?
 
  • #16
varunika said:
so... back to the asked question..
in the query.. we have to find the time wen energy at node=energy at anti node.
KE of node = PE of anti node
∴ 1/2Δmv2 = 1/2ΔmV(dy/dx)2
v=λf and f=1/T
∴ v=λ/T
would i be wrong if i relate this in the equation... since we must need to relate the timeperiod with energy?
and my other doubt is... that equation of wave is not given... so how to find dy/dx and λ?

The formula for a standing wave is just the product of two sinusoidal functions.

y = A sin(kx + δ) sin(ωt + ø),

Where δ and ø are just phases
 
  • #17
well... i seem to be going no where towards solving my question :(
 
Back
Top