About velocity equation in simple harmonic motion

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Homework Help Overview

The discussion revolves around understanding the velocity equation in the context of simple harmonic motion, specifically focusing on the components of velocity derived from circular motion. Participants are exploring the relationship between angular displacement and linear velocity components.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to understand the derivation of the velocity component equations, particularly questioning the presence of a negative sign in the expression for Vx. They are also discussing the angles involved in the velocity vector and how they relate to the coordinate system.

Discussion Status

The discussion is active, with participants providing insights and clarifications regarding the geometric interpretation of the problem. Some have offered guidance on visualizing the angles and components involved, while others are still grappling with the implications of the signs in the equations.

Contextual Notes

There is a note from the original poster indicating a preference for explanations that do not involve derivatives, which may influence the direction of the discussion. Participants are also referencing diagrams to clarify their points, suggesting a reliance on visual aids for understanding the concepts.

Thaurron
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Homework Statement


Actually this is not a problem. This is about something i couldn't understand.

Y5y45A.jpg


Homework Equations


dairesel3.jpg

where wt equals θ

The Attempt at a Solution


I can understand that x=r.cosθ it is obvious. But i can't understand why Vx=-V.sin(w.t) When i take a triangle from that circle i end up finding Vx equal to V.sin(w.t) . Where do we get that minus? Also how does he find the angle as 90+wt in that equation? I think it should be 90 - wt

Note: I don't want the explanation with derivative.
 
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Thaurron said:
how does he find the angle as 90+wt in that equation?
It's all there in the diagram. The angle from the +ve X axis anticlockwise around to the velocity vector is shown as 90+theta. If that's not clear, project the blue radius line through the particle to cut the angle into two parts.
 
Ummm.. I think i couldn't express my problem. I can understand why is that angle is 90 + theta. My problem is different.

If you take triangle which has a pink vector leftward and green vector which making an angle of alpha with the pink vector (where alpha is 90 - theta) and put the Vy component into the emptry section of that triangle you will see that Vx equals V.cos(alpha) which is 90 - theta so equation becomes Vx=V.cos(90-theta)

As i know, that 90 degree in the cosinus will turn it into a sinus but won't add a minus to it since 90 - theta is in the first section of coordinate system.

Summary: My problem is with that minus... Btw thank you for helping.
 
Thaurron said:
If you take triangle which has a pink vector leftward and green vector which making an angle of alpha with the pink vector (where alpha is 90 - theta) and put the Vy component into the emptry section of that triangle you will see that Vx equals V.cos(alpha) which is 90 - theta so equation becomes Vx=V.cos(90-theta)
But in that diagram you have drawn Vx as a left-pointing vector. Since right is positive, you should take Vx as right-pointing. You then see that it is Vcos(90+theta), hence the minus sign.
 
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Got it. Thanks a lot.
 

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