Graphs relating to simple harmonic motion

  • #1
RoboNerd
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Homework Statement



untitled1.jpg
Untitled 2.jpg

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!
 

Answers and Replies

  • #2
TSny
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Please explain how you got your answer.
 
  • #3
Physics-Tutor
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Velocity can also be negative...
 
  • #4
haruspex
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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.
 
  • #5
RoboNerd
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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?
 
  • #6
haruspex
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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?
 
  • #7
RoboNerd
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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
 
  • #8
haruspex
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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.)
 

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