TBR: Spring Oscillation Question

[justadream]

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TBR Physics page 244

The mass is initially pulled to the RIGHT. I understand why B represents the velocity vs. time graph.

I don't understand why choice C represents the acceleration vs. time graph (TBR asserts this in the answer explanation). The mass is displaced to the right initially. Thus, shouldn't the acceleration initially be negative (as in, choice D)?

The choices in order are: B, C, D

 

[justadream]

bump!

 

[The Brown Knight]

yeah, doesn't make sense to me either. even if you just look at the tangent line to the velocity graph, it starts out negative.
may be a typo? they seemed to include it secondarily to the answer

 

[techfan]

The acceleration is what drives the restoring force (I.e. It's what pulls the spring back when it gets displaced) so it has to be going in the opposite direction of the velocity (or out of phase {fill in the rest of the graph to make round objects}). Since the velocity after you release it is going to the left the acceleration has to be going to the right otherwise the spring would just keep going to the left and nothing would pull it back.

 

[justadream]

@techfan

Doesn't acceleration only reverse once you pass the equilibrium point?

If what you are saying is correct, then velocity must be maximized the INSTANT the object leaves its max displacement. But velocity starts from 0 and increases until it reaches it's max at the displacement = 0 position.

 

[techfan]

I'm totally wrong, for some reason (late night maybe) I saw that velocity started at a maximum going left, but it starts from 0.

If you look at it in terms of derivatives, x(t)=A*cos(ωt) where A= the amplitude (scaling factor, or how loud something is) ω is angular velocity and t=time

v(t)=dx(t)/dt=x'(t)=-Aωsin(ωt) (derivative is change or slope of function and derivative of cos(x) is -sin(x) {if there is a constant in the function <in this case ω> you just multiply the entire function by that constant})

a(t)=dv(t)/d(t)=v'(t)=x''(t)=-Aω^2*cos(ωt) (derivative of sin(x) is cos(x) so derivative of -sin(x) is -cos(x))

Anyways, the important parts are the -sin and -cos parts. Sin(0)=0 and cos(0)=1, but this is negative sin and negative cos so it's (like you said) 0, and -1. So yeah, D is the acceleration, and C is displacement (x(t)=A*cos(ωt))
Sorry about that.

 

[type12]

Velocity is max at zero displacement. Imagine the object from the beginning (being held to the right):

The force is pushing to the left, so acceleration is to the left. As the distance from the resting length approaches zero, the force (and therefore acceleration) goes to zero. Bear in mind, the object has been accelerating the whole time, just to a less and less degree. Now, when it passes the resting length, the force is in the opposite direction to the velocity. Therefore, it is decelerating. You get that deceleration until you reach the end point, where the force is maximized, but the velocity has reached zero (because it's about to change direction).

That entire story is embedded in the first trough on the graph, from the first zero y displacement to the next zero y displacement. Repeat for the next part, and voila, crest!

 

[justadream]

@type12

So would you agree that choice D best represents the acceleration?

 

[type12]

@type12

So would you agree that choice D best represents the acceleration?

Yep