# Harmonic oscillator

1. Aug 12, 2012

### induing

hello,

new here and confused about newton second Law.

given:
vertical mass damper system, position of the mass: x(t)=sin(t)
velocity is: v(t)=cos(t)
acceleration is: a(t)=-sin(t)

function x(t): above x-axis describes position of the mass below the vertical equilibrium point, wich (below) is the positive direction of vector x

suppose I look at the movement between t=0 and t=T/4: when the mass is below the vertical equilibrium line and is moving to the ground

when I apply Newton second law (ma(t)) the vector a(t) must point downward because mass is going down, but when I look at the the function a(t) is it negative, wich means the vector a(t) is pointing upward (ponting in negative direction)...

I'm off the right track, but don't see my fault..

so if anyone can help, please?

grtz

2. Aug 12, 2012

### CAF123

If I understand this correctly, the acceleration vector should point upwards because, on the way down, the mass is approaching its lowest position and will have zero velocity there. This upward acceleration vector provides the necessary deceleration of the mass. The velocity vector points downward, however.

3. Aug 12, 2012

### induing

so the term (ma(t)) isn't the vector wich is related to the movement of the mass?

i find that rather weird, because ma(t) equals the vector sum of all the (external) forces, so i would pressume that this net force gives the direction...

and if I do understand this correctly, I cannot predict the movement of the mass just by looking at the acceleration vector?

grtz

4. Aug 13, 2012

### CAF123

The movement of a body is generally represented by the velocity vector, which is always tangential to the motion. For example, if a car was travelling eastwards, the velocity vector would point eastwards.

The acceleration vector, however, acts in the direction opposite to motion if the object is slowing down. In the above example, if the car was travelling eastwards and slowed down to turn a corner,say, the velocity vector points eastwards, but the acceleration vector points westwards. We say the car has a negative acceleration (in my first post deceleration should be replaced with negative acceleration to be more precise). However, in accelerating a car, for example, the velocity and acceleration vectors do point in the same direction. (net force on car is in this direction). $$\sum F_x = ma_x$$

5. Aug 14, 2012

### induing

thanks for the clear and brief explanation!
must have been very confused: force wich is needed to give a body certain amount of movement, is indeed the thing wich control way the way the body change it's movement

more friction will slow it down down, so decceleration kicks in, but body off course is still moving some time in the same direction.

thank you big time,sir!
grtz

6. Aug 15, 2012

### CAF123

Perhaps to give another example, have you ever looked at circular motion? The velocity vector is always tangent to the circle if a body is undergoing circular motion. But the acceleration vector can be in quite unexpected places. If there is no tangential acceleration (that is, constant velocity and uniform circular motion) then the acceleration vector is radially inwards (the body experiences a centripetal acceleration). However, if there is a tangential acceleration, the resultant acceleration vector is the vector sum of the centripetal acc. and tangential acc. and we get a vector that is slightly offset from being directed exactly towards the centre of the circle.

7. Aug 15, 2012

### induing

yeah I know the dynamics of circular motion, constant velocity or accelerated
the normal (radial) component is needed to change movement of direction and is always presented

if you want a accelerated circular movement (speeding up in a curved road) the tangential vector wich produces change in velocity kicks in

acceleration vector then consists of 2 vector.

have been confused with force direction as being the direction of the acceration vector, wich just indicates the way the speed is changing an not the movement itself

soon examination of dynamics,
so many thanks again for the quick reply!

grtz