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Simple harmonic motion velocity & time

  • #1
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Homework Statement



The velocity of an object in simple harmonic motion is given by v(t)= -(0.269 m/s)sin(15.0t + 2.00π), where t is in seconds. What is the first time after t=0.00 s at which the velocity is -0.150 m/s?


Homework Equations


already given in question


The Attempt at a Solution


so i tried finding the time by simply plugging -0.150 m/s in the equation v(t)= -(0.269 m/s)sin(15.0t + 2.00π) and solved for time i got a negative answer..when i tried to put that in it was wrong and so was the positive can someone tell me what i am doing wrong thankx :))
 

Answers and Replies

  • #2
CompuChip
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Can you show us your work?
Probably you are almost there. Note that the question explicitly asks for the first solution after t = 0. Do you understand what this has to do with the given equation which contains a sine?
 
  • #3
there are lots of different solutions to the inverse sine of a number, though sine is not directly symmetrical about the y axis.

if you're still struggling, plot a rough graph of v against t.

the question is asking for the first time V = -0.15 where t is posative.


oh, and you're using radians?
 
  • #4
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this is how i did it the long version

v(t)= -(0.269 m/s)sin(15.0t + 2.00π)
-0.15= -(0.269 m/s)sin(15.0t + 2.00π)
-0.15/-0.269 = sin (15.0t + 2.00π)
[sin-1(-0.15/-0.269)]-2.00π =15 t

-5.691668/15 = t
-0.379 = t
 
  • #5
138
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this is how i did it the long version

v(t)= -(0.269 m/s)sin(15.0t + 2.00π)
-0.15= -(0.269 m/s)sin(15.0t + 2.00π)
-0.15/-0.269 = sin (15.0t + 2.00π)
[sin-1(-0.15/-0.269)]-2.00π =15 t

-5.691668/15 = t
-0.379 = t

CompuChip can you clarify what you meant thanks ..and yes I am using radians all the way through
 
  • #6
CompuChip
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What you did seems correct, except it doesn't give you the answer you want. Remember that the sine function is periodic, so it keeps repeating itself. You can easily see this if you make a graph. Suppose that the period is P (for sin(t) the period P = 2pi for example, while for sin(3t) it is P = 2pi/3). If you have a solution to your equation you can add or subtract any multiple of P (like 2P, 12P, 12356P) and get another solution.

So first you should find out the period of sin(15.0t+2pi). Then add an appropriate multiple until you get something > 0.

Sorry I don't have much time to explain thoroughly right now, is it clear what I'm talking about?
 
  • #7
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i'm a bit confused so i know what my angular frequency is which is 15 rad/s and the period of that is 2pi(15)...... and then i get confused are you saying that i should change the period and make it 3pi(15) and plug that in to the eqaution to get the answer??
 
  • #8
diazona
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You're right that the angular frequency is 15 rad/s, but the period is not 2π(15).
for sin(t) the period P = 2pi for example, while for sin(3t) it is P = 2pi/3
Do you see the pattern in that? Once you do, what is the period of sin(15t)?
If you have a solution to your equation you can add or subtract any multiple of P (like 2P, 12P, 12356P) and get another solution.
Once you have found the period correctly, just follow that instruction. You can take the solution you found and add or subtract any multiple of P to get another solution; in fact, you can create an infinite list of solutions this way. (Don't try to write out the whole infinite list! ;-p Just figure out enough that you can see the pattern.) Now, out of all these solutions, what is the smallest one that is greater than 0?
 
  • #9
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would the period be then 2pi/15??
 
  • #10
CompuChip
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Yes, very good.

By the way, did you also plot the graph as I suggested? (Using a graphical calculator or a computer program, for example)
 
  • #11
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i have another question i was wondering where this period would go in the equation of velocity...also the nxt part has to do with position and was wondering if this looked right
v(t)= -(0.269 m/s)sin(15.0t + 2.00π) changes to
x(t)= (0.269/15) cos (15.0t+2.00π)??/
 
  • #12
CompuChip
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That looks very much ok to me.

If you doubt yourself, remember that the velocity is the derivative of x(t) with respect to time. So you can take the derivative and check that you do get v(t).
 
  • #13
138
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also with my new period how do i use it with the velocity equation??
 
  • #14
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hi ok so i finally understood what to do and was wondering if you could verify my answer since they removed the question from my homework application ...i got 0.0399 seconds
 
  • #15
CompuChip
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In 4 decimals, I got 0.0394 but that's probably just a rounding difference (you might have rounded an intermediate result where I didn't).
So you solved the question, very well!
 

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