About velocity equation in simple harmonic motion

AI Thread Summary
The discussion centers on understanding the velocity equation in simple harmonic motion, specifically why Vx is expressed as -V.sin(wt) instead of V.sin(wt). The confusion arises from the interpretation of angles in the context of a triangle formed by the velocity vector and the radius in circular motion. It is clarified that the angle from the positive X-axis to the velocity vector is 90 + theta, leading to the negative sign in the equation. The participants emphasize the importance of the direction of the vectors in determining the sign of Vx. Ultimately, the correct interpretation resolves the initial confusion regarding the negative sign in the velocity equation.
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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