Instantaneous Particle Velocity from Wave

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SUMMARY

The discussion focuses on calculating the instantaneous particle velocity from the wave equation D(x,t) = Asin(kx-ωt+ϕ). The derivative of displacement, D'(x,t) = -ωAcos(kx-ωt+ϕ), is used to find the velocity, resulting in the expression -18.84cos((π/15)x-3πt). The challenge lies in the lack of specific values for time or distance, which are necessary for numerical evaluation. The solution suggests using known position values to determine the phase and facilitate the calculation of velocity.

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  • Familiarity with trigonometric functions and their applications in physics
  • Ability to interpret phase relationships in wave motion
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rocapp
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Homework Statement


See the attached picture.



Homework Equations


D(x,t) = Asin(kx-ωt+ϕ)


The Attempt at a Solution


The instantaneous velocity of a particle would be the derivative of displacement:
D'(x,t) = -ωAcos(kx-tω+ϕ)

Then plug in the known values:
= -18.84cos((π/15)x-3πt)

The above should be the answer, but since I do not have a specified value for time or distance, I am not sure how to come to a numerical answer. Thanks in advance!
 

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rocapp said:
since I do not have a specified value for time or distance
You have an indirect one: At point 1, the position is 0. This allows to determine a value for the phase (kx-ωt+ϕ), which you can use in the equation for the velocity.
 
Thanks!
 

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