Graphs relating to simple harmonic motion

[RoboNerd]

Homework Statement

Homework Equations

None.

The Attempt at a Solution

Hi everyone. Apparently 5 is the right answer, although I chose D.

Could anyone please weigh in with their thoughts about why 5 is right and my answer is apparently wrong?

Thanks!

 

[TSny]

Please explain how you got your answer.

 

[Physics-Tutor]

Velocity can also be negative...

 

[haruspex]

Velocity can also be negative...

True, but that does not rule out any of the graphs. There is no claim that the graph represents an entire cycle. There is a better reason for choosing 5. What would the graph of velocity actually look like?

Edit: please do not post an answer to that on this thread, at least not until RoboNerd has had a chance to answer it.

 

[RoboNerd]

Hi everyone, robonerd is back, of course.

I know that at xmin and max, the velocity [with kinetic energy] = 0 instantaneously. Thus, I narrow down to B and D.
However, I know that with a variable spring force giving a variable acceleration, I will not have the velocity changing in a linear manner [constant acceleration with constant slope], so I rule out B. D is thus a potential answer. Why is D wrong?

 

[haruspex]

Hi everyone, robonerd is back, of course.

I know that at xmin and max, the velocity [with kinetic energy] = 0 instantaneously. Thus, I narrow down to B and D.
However, I know that with a variable spring force giving a variable acceleration, I will not have the velocity changing in a linear manner [constant acceleration with constant slope], so I rule out B. D is thus a potential answer. Why is D wrong?

Can you write an equation relating velocity and x?

 

[RoboNerd]

Yes. Using conservation of energy I have:

( 1 / 2 ) * k * A^2 = a constant value = (1 / 2) * m * v^2 + ( 1/ 2) * k * x^2

 

[haruspex]

Yes. Using conservation of energy I have:

( 1 / 2 ) * k * A^2 = a constant value = (1 / 2) * m * v^2 + ( 1/ 2) * k * x^2

Good. Can you recognise that form as a common shape? (Think of v as the y coordinate.)