Simple Harmonic motion help neededWatch

Announcements
#1
Ok so my textbook says:

Acceleration is always in the opposite direction to displacement.

How can this be the case when at max displacement acceleration= max negative but the particle has max velocity at the equilibrium position so as it moves towards its equilibrium position from max displacement it must accelerate from 0 -> max m/s therefore acceleration is in the same direction to the displacement (both acting towards equilibrium)?

Please tell me where I have gone wrong
0
5 years ago
#2
Acceleration is the change in velocity.
A particle hits max negative acceleration when the particle changes direction (turns back towards the equilibrium position at max displacement), so that part is fine.

When the particle is heading towards 0 displacement, there still is some acceleration, which causes the increase in speed, but the acceleration will decrease to 0 at the same time displacement does. Remember that the size of the acceleration is getting smaller as the particle approaches the equilibrium position, but as long as there is some acceleration, there will be a change in speed. When the particle moves past the equilibrium position, it'll start accelerating towards 0 displacement again, making the particle slow down until it reaches max displacement again.

There's probably a much better way to explain this, but my textbook doesn't do a great job either.

Short version:
Acceleration gets smaller as displacement approaches 0, but since there is some acceleration, the particle speeds up. It just doesn't speed up as quickly when displacement gets smaller.
0
5 years ago
#3
(Original post by Davelittle)
Ok so my textbook says:

Acceleration is always in the opposite direction to displacement.

How can this be the case when at max displacement acceleration= max negative but the particle has max velocity at the equilibrium position so as it moves towards its equilibrium position from max displacement it must accelerate from 0 -> max m/s therefore acceleration is in the same direction to the displacement (both acting towards equilibrium)?

Please tell me where I have gone wrong
I think I understand what you're asking but correct me if I'm wrong, you want to know about the direction of the acceleration as the bob heads back to equilibrium because it's accelerating, obviously, because the velocity goes from zero to maximum? I don't know exactly what you mean but this animation might help? http://en.wikipedia.org/wiki/File:Os...g_pendulum.gif If it's something you're struggling to visualise.
0
5 years ago
#4
(Original post by Davelittle)
Ok so my textbook says:

Acceleration is always in the opposite direction to displacement.

How can this be the case when at max displacement acceleration= max negative but the particle has max velocity at the equilibrium position so as it moves towards its equilibrium position from max displacement it must accelerate from 0 -> max m/s therefore acceleration is in the same direction to the displacement (both acting towards equilibrium)?

Please tell me where I have gone wrong
Displacement is zero at equilibrium, so it is measured outwards from there.
Displacement gets bigger (more positive) as you move further away from equilibrium.
The acceleration points in the opposite direction, towards the equilibrium position, because the (restoring) force is pointing in that direction.
So displacement is measure outwards away from the equilibrium point, while acceleration has a direction inwards towards the equilibrium point. Hence the acceleration is negative in sign.
0
#5
But as it is going from max displacement -> equilibrium I don't understand how the acceleration is acting in the opposite direction (would this decelerate it instead of accelerate it to max v at equilibrium)?
0
5 years ago
#6
The acceleration ALWAYS acts towards the equilibrium point hence why the object has maximum velocity at equilibrium as it is being accelerated in that direction. When the object passes through equilibrium then the acceleration is acting towards the centre still, which is in the opposite direction to the direction the object is moving in hence why it slows down as it approaches maximum displacement or amplitude.
1
5 years ago
#7
(Original post by Davelittle)
But as it is going from max displacement -> equilibrium I don't understand how the acceleration is acting in the opposite direction (would this decelerate it instead of accelerate it to max v at equilibrium)?
I think you're confusing deceleration with slowing down, the 2 don't always mean the same thing.

EG: If I accelerate something from rest to 5 m/s in 10 seconds, then accelerate it to 6 m/s in another minute, the acceleration has decreased but the speed is still increasing, it's just not increasing as quickly as it did when the acceleration was bigger.

When something is already accelerating, deceleration will first slow down how quickly it's speed is increasing, before the velocity starts to decrease. Provided the thing was accelerating in the first place.
0
#8
I think I understand it now, yay! thanks to everyone.
0
X

new posts
Latest
My Feed

Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

University open days

• London Metropolitan University
Tue, 21 May '19
• Brunel University London
Wed, 22 May '19
• University of Roehampton
Wed, 22 May '19

Poll

Join the discussion

How did your AQA A-level Economics Paper 1 go?

Loved the paper - Feeling positive (51)
14.21%
The paper was reasonable (179)
49.86%
Not feeling great about that exam... (88)
24.51%
It was TERRIBLE (41)
11.42%