SHM Troubles (1 Viewer)

.ben

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7. Given a=-16x and that when x=3 and v=16, t=0, find a relation between x and t.

16. A particle is moving in SHM starts from rest at x=5 and after 2 seconds reaches x=2.5. Find an expression for the displacement at time t>_0.

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Riviet

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7. Using the fact that a=d(v2/2)dx=-16x, integrate with respect to x to find an expression with v and x. Then use v=dx/dt and integrate with respect to x OR t (the question didn't specify in terms of x or t). Don't forget to find the constant each time you integrate.
 
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vafa

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.ben said:
7. Given a=-16x and that when x=3 and v=16, t=0, find a relation between x and t.

16. A particle is moving in SHM starts from rest at x=5 and after 2 seconds reaches x=2.5. Find an expression for the displacement at time t>_0.

Thank you.
this is the solution to question7

View attachment 13597
 
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vafa

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.ben said:
7. Given a=-16x and that when x=3 and v=16, t=0, find a relation between x and t.

16. A particle is moving in SHM starts from rest at x=5 and after 2 seconds reaches x=2.5. Find an expression for the displacement at time t>_0.

Thank you.
This is the solution to question 16 in two pages:

View attachment 13602

View attachment 13603
 
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vafa

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An Alternative method:
question16.

If a particle moves in a simple harmonic motion, then the equation of its displacement is one of these forms:
x=acos(nt+alpha) or
x=asin(nt+alpha) or
x=bsin(nt)+ccos(nt)
but at t=0, x=5 so you need to think about cosine not sine since cos0=1
x=acos(nt+alpha) (1)
5=acos(alpha)

v=-ansin(nt+alpha)
but v=0 so sin(alpha)=0 so alpha=0
hence x=acos(nt)
x=5 when t=o so a=5
so x=5cos(nt)
x=2.5 when t=2
cos(2n)=1/2
2n=pi/3
n=pi/6

eventually the equation of displacement becomes:
x=5cos(pi/6 t)
 

.ben

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vafa said:
but at t=0, x=5 so you need to think about cosine not sine since cos0=1
I don't get this bit? Why do you have to use cosine? Arn'te thaey both the same?:confused:
 

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