Resolution of Vectors

Thread Starter

logearav

Joined Aug 19, 2011
248
Revered members, i have attached two images which depicts obtaining an expression for velocity and acceleration of particle executing SHM.

1)In the first attachment, velocity component vcosΘ is parallel to vertical diameter of the circle and vsinΘ is perpendicular to the vertical diameter.
In the second attachment, acceleration component vcosΘ is perpendicular to vertical diameter and vsinΘ is parallel to vertical diameter.
2)I know perpendicular component is sine and parallel component is cosine. But i cant understand why perpendicular component to vertical diameter is taken as cosine in acceleration expression and parallel component as sine? Please help revered members, with regard to guidelines for resolving vectors in cosine and sine components
 

Attachments

Thread Starter

logearav

Joined Aug 19, 2011
248
My doubt is, the perpendicular line PN is taken as sine component for velocity measurement, but the same perpendicular line PN is taken as cosine component(second attachment) for acceleration. Why?
 

BillO

Joined Nov 24, 2008
992
Acceleration in a particular direction is at a minimum when the velocity in that direction is at a maximum, and vice-verse. The is true in this case as well. Once P reaches Y, its velocity parallel to the X axis will be at a maximum, therefore its acceleration in that direction will be 0.
 

Thread Starter

logearav

Joined Aug 19, 2011
248
Acceleration in a particular direction is at a minimum when the velocity in that direction is at a maximum, and vice-verse. The is true in this case as well.
Once P reaches Y, its velocity parallel to the X axis will be at a maximum, therefore its acceleration in that direction will be 0.
But why acceleration is zero, when velocity is maximum?
 

BillO

Joined Nov 24, 2008
992
Acceleration is the rate of change of velocity. If velocity is at a maximum, even for a very brief time, it is not changing, therefore acceleration is zero.
 

Thread Starter

logearav

Joined Aug 19, 2011
248
Thanks for the reply BillO. Since velocity is maximum parallel to X axis, we put sine component and at the same time since acceleration is minimum parallel to X axis, we put cosine component. But why sine for maximum and cosine for minimum?
 

studiot

Joined Nov 9, 2007
5,003
You said yourself that the velocity components are vcosθ and vsinθ.

What happens if you differentiate a sin or cos function?

(Have you done this in maths?)
 

Thread Starter

logearav

Joined Aug 19, 2011
248
Thanks for the reply studiot. My question is why should we choose, vsin(thetha) for maximum velocity and vcos(thetha) for minimum velocity. Why not vcos for maximum and vsin for minimum
 

BillO

Joined Nov 24, 2008
992
Well you could, but you'd have to use a different angle. Given the situation as described, and the angle (θ) as defined, vcos(θ) is the correct model for the vertical component of v, and vsin(θ) is the correct horizontal component of v.

For instance, when p reaches Y, θ=pi/2, thefore sin(θ)=1 and cos(θ)=0. From inspection we can see this means the velocity is parallel to X, X', which is indeed the case.
 
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Thread Starter

logearav

Joined Aug 19, 2011
248
You said yourself that the velocity components are vcosθ and vsinθ.

What happens if you differentiate a sin or cos function?

(Have you done this in maths?)
differentiation of sine is cosine and differentiation of cosine is -sine.
 

studiot

Joined Nov 9, 2007
5,003
Remember that we use an arm rotating anticlockwise from the x axis to represent angles.
If r is the length of the arm, the projection of this arm on the x axis is rcosθ.
ie x=rcosθ
The projection on the y axis is rsinθ ie y=rsinθ

Historically this is how sinusoidal waves generated by harmonic oscillators was linked to the rotating arm.

attached is the best explanation I know of of why we use cos and sine etc as we do.

The reason I asked about calculus was because

if x = rcosθ

then differentiating this with respect to time yields velocity

dx/dt = d(rcosθ)/dt = -rsinθ

differentiating again with respect to time yields acceleration.

Now perhaps yopu know enough calculus to find out where the max and minima for velocity and acceleration are. If you do you will see mathematically why the max velocity occurs at the min accel and vice versa.
 

Attachments

Thread Starter

logearav

Joined Aug 19, 2011
248
Once P reaches Y, its velocity parallel to the X axis will be at a maximum, therefore its acceleration in that direction will be 0.
Velocity in SHM is (√(a^2 - y^2)w
BillO,
So when the particle P is at O, that is at mean position y = 0 and velocity is maximum i.e v =±aw
When the particle P is at Y, the particle is at extreme position i.e y = ±a, so velocity is zero.
My calculation is contradiction to your statement. Sorry, i might be completely misinterpreting your statement. Just to clear my doubt, i asked.
 
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Thread Starter

logearav

Joined Aug 19, 2011
248
studiot,
Thanks for your detailed reply. But my question is , in what way differentiation of sine leading to cosine and differentiation of cosine leading to - sine influence the horizontal and vertical components?
 

studiot

Joined Nov 9, 2007
5,003
But my question is , in what way differentiation of sine leading to cosine and differentiation of cosine leading to - sine influence the horizontal and vertical components?
What do you mean? Please expand.
 

Thread Starter

logearav

Joined Aug 19, 2011
248
studiot,
x = rcosθ and y = rsinθ
dx/dt gives -rsinθ which is velocity and d/dt(dx/dt) gives -rcosθ which is acceleration.
Similarly differentiating y with respect to time, gives velocity as rcosθ and acceleration as -rsinθ. How this differentiation influences the resolution of parallel component as vsinθ and perpendicular component as vcosθ, wrt to my first attachment?
ii) Again, how this differential calculus helps us to determine where maximum velocity occurs at minimum velocity and vice versa?
You have taken pains to make things clearer to me and many thanks for that. But i need some more help.
 
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