About velocity equation in simple harmonic motion

In summary, the conversation discusses a problem with understanding the equation Vx=-V.sin(w.t) and how the angle of 90+wt is found in the equation. The person asking the question believes it should be 90-wt. However, after further explanation and using a triangle to demonstrate, it is clarified that the equation should be Vx=V.cos(90+theta) and the minus sign is due to the direction of the vector. The issue is resolved and the person thanks the other participant for their help.
  • #1
Thaurron
3
0

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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  • #2
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.
 
  • #3
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.
 
  • #4
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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Likes Thaurron
  • #5
Got it. Thanks a lot.
 

1. What is the equation for velocity in simple harmonic motion?

The equation for velocity in simple harmonic motion is v = ω√(A^2 - x^2), where v is velocity, ω is the angular frequency, A is the amplitude, and x is the displacement from equilibrium.

2. How does the velocity equation change if the object is at equilibrium?

If the object is at equilibrium, the displacement (x) is equal to 0, so the velocity equation simplifies to v = ωA. This means that the velocity is directly proportional to the angular frequency and amplitude.

3. Can the velocity in simple harmonic motion ever be negative?

Yes, the velocity in simple harmonic motion can be negative. This occurs when the object is moving in the opposite direction of its equilibrium position. For example, if the object is at its maximum displacement to the right, its velocity will be negative as it moves back towards equilibrium.

4. How does the mass of the object affect the velocity in simple harmonic motion?

The mass of the object does not directly affect the velocity in simple harmonic motion. However, it does affect the angular frequency (ω) which is used in the velocity equation. A larger mass will result in a smaller ω, and therefore a smaller maximum velocity.

5. What is the relationship between velocity and acceleration in simple harmonic motion?

In simple harmonic motion, the acceleration and velocity are always perpendicular to each other. This means that the velocity is constantly changing, but the acceleration remains constant. Additionally, the maximum velocity occurs at the equilibrium position, where the acceleration is 0.

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