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Simple harmonic motion

  • Thread starter Matt512
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Homework Statement


A particle moves with simple harmonic motion in a straight line with amplitude 0.05 m and the period 12 s.Find (a) the maximum speed , (b) the maximum acceleration of the particle.Write down the values of the constants P and Q in the equation
x/m = P sin [Q)(t/s)]

Homework Equations




The Attempt at a Solution


Well i tried to find the speed using A w sin w t
but my problem is , how do i find the t ?
 

Answers and Replies

  • #2
DrClaude
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##t ## stays a variable. You should be able to restrict your analysis to one oscillation, as the motion will be periodic.
 
  • #3
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if i use the equation A w sin wt
i should replace t with what ?
 
  • #4
DrClaude
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You are asked to calculate the maximum speed and acceleration. Do you know how to do that? For that, you need to be able to write position as a function of time. You then need to find the times that will maximize each of these values, for which, as I hinted, you can restrict yourself to one oscillation.

By the way, the questions asks you to write it as x/m = P sin [Q(t/s)]. Use can use this instead of A and ω. Have you found P and Q?
 
  • #5
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You are asked to calculate the maximum speed and acceleration. Do you know how to do that? For that, you need to be able to write position as a function of time. You then need to find the times that will maximize each of these values, for which, as I hinted, you can restrict yourself to one oscillation.

By the way, the questions asks you to write it as x/m = P sin [Q(t/s)]. Use can use this instead of A and ω. Have you found P and Q?
Its still not clear in my head :( Can you give me an example of how to write position as a function of time and where to take the max speed ?
I am still trying to do the part (a) , when i finish i will do the remaining parts
 
  • #6
DrClaude
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Can you give me an example of how to write position as a function of time
An example is given in the question:
$$
x = P \sin(Q t)
$$
You have position on the left-hand side, and a function of time on the right-hand side. Therefore, you can say that position is a function of time, that it varies with time.

I am still trying to do the part (a) , when i finish i will do the remaining parts
Finding P and Q is the first thing to do. I don't see how you can answer (a) or (b) without it. It may be the way the question is written that is confusing you. I read the last sentence of the problem as a hint, rather than a part (c) to be answered.
 
  • #7
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Ah i see
Yeah the question was confusing me xD

Well
x=Psin(Qt) you compare it to
x=Asinwt

p will be equal to 0.05
and w(Q) will be π/6 right ?
 
  • #8
DrClaude
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Ah i see
Yeah the question was confusing me xD

Well
x=Psin(Qt) you compare it to
x=Asinwt

p will be equal to 0.05
and w(Q) will be π/6 right ?
Correct.
 
  • #9
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Correct.
I have 1 question
When do you know that you have to use either A w cos wt or A w sin wt ?
 
  • #10
DrClaude
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I have 1 question
When do you know that you have to use either A w cos wt or A w sin wt ?
It depends on the initial conditions (and your choice of ##t=0##). For a spring that is stretched at ##t=0## and then let go, the cosine will give you the right equation as you will get ##x(0) = A##.
 
  • #11
DrClaude
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I should probably add that the general solution to the equation of motion of an harmonic oscillator is
$$
x(t) = A \cos (\omega t) + B \sin (\omega t)
$$
where ##A## and ##B## are set by the initial conditions (position and velocity at ##t=0##). It is left as an exercise :wink: to show that this can be rewritten as
$$
x(t) = C \sin (\omega t + \phi)
$$
 
  • #12
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Well if i choose X= A sin wt

then v=dx/dt = A w cos wt which is what i want

if i choose x = A cos wt it would be the inverse
right ?
 
  • #13
DrClaude
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Well if i choose X= A sin wt

then v=dx/dt = A w cos wt which is what i want

if i choose x = A cos wt it would be the inverse
right ?
Yes, and this corresponds, as I mentioned above, to the arbitrary choice of the zero of time.
 
  • #14
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Well using v=dx/dt = A w cos wt i get 0.026 ms
What formulae should i use for the acceleration ? Is it the same principle as for the speed ?
 
  • #15
DrClaude
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Well using v=dx/dt = A w cos wt i get 0.026 ms
What formulae should i use for the acceleration ? Is it the same principle as for the speed ?
Yes, same principle.
 
  • #16
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so its A w^2 cos wt ?
 
  • #17
DrClaude
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so its A w^2 cos wt ?
Yes, that's the equation for acceleration.

[Edit: but with a minus sign. Thanks to BvU for spotting this.]
 
Last edited:
  • #18
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Thanks very much for you help :)
 
  • #19
BvU
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so its A w^2 cos wt ?
Actually, there should appear a minus sign when differentiating a cosine.
I hope that it is clear to you that when you start with
X= A sin wt

then v=dx/dt = A w cos wt
the acceleration is given by a = dv/dt = -A w2 sin(wt)

And conversely, when you start with
X= A cos wt
then v=dx/dt = -A w sin(wt)
and a = dv/dt = -A w2 cos(wt)​
 
  • #20
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Actually, there should appear a minus sign when differentiating a cosine.
I hope that it is clear to you that when you start with the acceleration is given by a = dv/dt = -A w2 sin(wt)

And conversely, when you start with
X= A cos wt
then v=dx/dt = -A w sin(wt)
and a = dv/dt = -A w2 cos(wt)​
well for v and a i got the correct answer by using the cos equation
then when you derive the equation one should be cos and the other sin , for the v and a right?
and thd minus sign is it necessary to put??
 
  • #21
DrClaude
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well for v and a i got the correct answer by using the cos equation
then when you derive the equation one should be cos and the other sin , for the v and a right?
and thd minus sign is it necessary to put??
The minus sign is necessary to get the right time dependence for acceleration, ##a(t)##. The problem asks for the maximum acceleration, which implicitly means the maximum in absolute value, so the answer will be positive.
 
  • #22
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Thanks :)
 

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