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Simple Harmonic Motion, Velocity at a certain time

  1. Nov 9, 2009 #1
    1. The problem statement, all variables and given/known data
    The velocity of an object in simple harmonic motion is given by v(t)= -(0.347 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 -0.100 m/s?

    2. Relevant equations

    3. The attempt at a solution
    -0.1 = -(0.347)sin(15t + 2[tex]\pi[/tex])
    0.2882 = sin(15t + 2[tex]\pi[/tex])
    15t = arcsin(0.2882) - 2[tex]\pi[/tex]
    t = -0.39939 s

    - its a neg time so i thought just adding 2[tex]\pi[/tex] would take me to the same point on graph but the positive side : 0.4284 s <-- still the wrong answer

    so im not too sure where im going wrong.. maybe its just my algebra or is it the whole approach to the question?
    thanks!
     
  2. jcsd
  3. Nov 9, 2009 #2
    can some one just confirm my algebra, because im pretty sure it is just a calculation error....? greatly appreciated thanks!!!
     
  4. Nov 9, 2009 #3

    duo

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    the sine function is periodic! what does that say about the 2 pi?

    0.2882 is correct.
     
  5. Nov 9, 2009 #4
    1) you say that its v(t) = -vmax sin(15t+2π) Why this 2π becomes 2^π?
    2) sin(0.297)= 0.288

    So 15t+2π= 2kπ + 0.297 or
    15t + 2π = 2kπ + π - 0.297

    3) I hope I am right about the numbers.. Your teacher REALLY hates you..
     
  6. Nov 9, 2009 #5
    so to get the right answer i had to add the negative value i got for the time to the period.
    T = 2[tex]\pi[/tex] / [tex]\omega[/tex] = 0.41888

    T + t = 0.0195 s ** and this is the right answer

    But can someone explain why I had to add it to one cycle (one period) ??
     
  7. Nov 10, 2009 #6
    Look inside your sine. It is 15t+2p. That means that for t=0 you have sin2p. That implies the body has already finished one period of motion which is not true ( or it is but you started measuring time after 1 period).

    think that sin(wt+a) (where a is an angle)

    when a =2p that means that before you started timing the body has already completed 1 period.
    That is the point of the angle there.
    Consider an a= 3p/2 that means that the initital position is not the equillibrium so in order to have the results wrt to x=o we needed that angle.

    Now Consider why you find the negative time.. 1) sin(15t+2p)=sin15t ( so you didnt have to evenr write 2p
    2) I cant understand why you use arcsin. Also when you have sin(15t+2p)= sin(0.297) <=>
    15t+2p= 2kp + 0.297 or 15t+2p=2kp+p-0.297.

    3) you need to add a period because you started the timing a period after it started moving. And even if you didnt the +2p implies it. It s better to get rid of it ( beacause you can ) and solve for t and the same result will pop up.

    Hope didnt confuse you more.
     
  8. Nov 10, 2009 #7

    Redbelly98

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    Staff Emeritus
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    Homework Helper

    Here is a simple example. Suppose you need to solve:

    sin(x) = 0.5​

    So of course we take the arcsin,

    x = arcsin(0.5) = π/6 (i.e. 30°)​

    And, of course, we can add integer multiples of 2π to get all possible solutions.

    But wait a minute ... 5π/6 (i.e. 150°) has a sine of 0.5 also, so that is another solution to the equation. But 5π/6 does not come from adding 2π to π/6, does it?

    So where does the extra solution 5π/6 come from? It is because there are two distinct angles that have a sine of 0.5.
     
  9. Dec 6, 2009 #8
    v(t)= -(0.347 m/s)sin(15.0t + 2.00π) is equal to v(t)= -(0.347 m/s)sin(15.0t). Now attempt to solve it in the same way as before and you will get t=0.0195s which is the correct answer.
     
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