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Free Oscillations of materials

  1. Dec 17, 2011 #1
    Problem

    A vertical oscillating system consist of three equal springs with elasticity coefficient c , among which a mass m=2kg is suspended.The system oscillates according to the following law of motion x(t)=0,4 cos 4t+0,5 sin 4t , (m) .

    Determine:

    1) The equivalent elasticity coefficient ce of reduced spring.
    2) For t=0 values for initial conditions x0 and v0
    3) The natural frequency k
    4) The oscillation period T
    5) The amplitude A


    Note = Please some help. I am high school student .Give me tips then i will try. Thanks to who helps.
     

    Attached Files:

  2. jcsd
  3. Dec 17, 2011 #2
    I got some ideas for the 3rd part of question.
    I have to find natural frequency. And for this need The natural period of the oscillation .
    And the natural period of the oscillation formulas in the attachment 2. (down)
    And the naturel frequency formula is it the attachment 1. (up)
    This are true or not ? And natural period of the oscillation means oscillation period ?
     

    Attached Files:

  4. Dec 17, 2011 #3
    1) The force by a spring is given as [itex]F = -kx[/itex], where [itex]x[/itex] is the displacement and [itex]k[/itex] is the elasticity coefficient, if you move your mass a small amount what is the net force on it from your three springs?
    2) You just need to evaluate x(0) and v(0), if you're stuck with the second see if you can remember how v is linked to x.
    3) There is a formula for the frequency of a spring given the elasticity coefficient and the mass.
    4) Period and frequency are very closely related, see if you can find the formula to convert them.
    5) The mass will be at its furthest points when v = 0.
     
  5. Dec 17, 2011 #4

    ehild

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    The equation of a simple harmonic motion is
    x(t) =Asin(ωt+θ),
    where ω is the angular frequency. It is 2pi times the natural frequency f :
    ω = 2p*f. The time period is the reciprocal of the frequency: T=1/f.
    A is the amplitude of the SHM and θ is a phase constant. x(t) is equivalent to the sum of a pair of sine and cosine functions of the same ωt:

    x(t) =Asin(ωt+θ)=(Acos(θ))sin(ωt)+(Asin(θ))cos(ωt)=asin(ωt)+bcos(ωt).

    The amplitude A is related to the coefficients of the sine and cosine terms on the right-hand side: A2=a2+b2.

    Compare the function in the problem with the function in the equation above. What is the frequency and period? What is the amplitude A of the equivalent single SHM?

    ehild
     
    Last edited: Dec 17, 2011
  6. Dec 17, 2011 #5
    Can you help whit numbers because my english very bad.Its not my main language :(
    x(t) =asin(ωt)+bsin(ωt) this is formul. And
    x(t)=0,4cos4t+0,5sin4t how can i compare .
    And note i am in high school.I thınk this kind of questions are hard for us ...I dont know where can i start.Please create equations whit my number .Thank you very much .
     
  7. Dec 17, 2011 #6

    ehild

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    Sorry, I meant x(t) =asin(ωt)+bcos(ωt).
    Your equation is x(t)=0.5sin(4t)+0.4cos(4t)

    a sin(ωt) ≡ 0.5 sin (4t)

    b cos(ωt) ≡ 0.4 cos(4t)

    Numbers and symbols of the same colour are identical.



    ehild
     
    Last edited: Dec 17, 2011
  8. Dec 17, 2011 #7
    I think i found some answers.in the attachments...Please can you check ? this are 3-4-5 part of my question.What about the first and second part.

    For second part. (For t=0 values for initial conditions x0 and v0 )

    x(0) = 0,4 cos (4x0)+0,5 sin (4x0) ===> Cos 0 = 1 and Sin 0 = 0
    x(0) = 0,4

    is this true ? and what does v0 means ?
     

    Attached Files:

  9. Dec 17, 2011 #8

    ehild

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    Your answers are correct.
    v0 is the velocity at time t=0. The velocity is the derivative of the displacement:
    Do you know what derivative is?
    ehild
     
  10. Dec 17, 2011 #9
    Yes i know a little.First derivative of displacement is velocity.And second derivative of displacement is acceleration.
    So İf t=0 then x0=0,4 ...And first derivative of x(t)=0,4cos4t+0,5sin4 is 0 isnt it ?
    Then Vo is equal 0 ? ( Derivative of constant number is 0 ) ? :S
     
    Last edited: Dec 17, 2011
  11. Dec 18, 2011 #10

    ehild

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    First do the derivation, then substitute t=0. Do you know the derivatives of cosine and sine?

    If derivation is not clear to you yet, you certainly was shown that the oscillatory motion can be imagined as projection of uniform circular motion. See picture. A point moves along a circle of radius R with angular velocity ω. At any angular position α, x=Rcos(α) and y=Rsin(α). The velocity V is tangent to the circle, and its magnitude is ωR. The components of the velocity vector are Vx=-Rωsin(α), Vy=Rωcos(α).

    It is a uniform circular motion, α=ωt. You can imagine that a function f(t)= Acos(ωt) is the x component of a circular motion along a circle of radius R=A, and the function g(t)=Asin(ωt) is the y component of the same circular motion. That is true also for the velocities, which are equal to the time derivatives of f and g.
    f'=Vf = -Aωsin(ωt) and g'=Vg=Aωcos(ωt).

    Your function is the sum of one sine and a cosine. The derivatives add up. Can you proceed?


    ehild
     

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    Last edited: Dec 18, 2011
  12. Dec 18, 2011 #11
    @ehild I am very happy to meet whit you.Thank you very much.You helped me and tried to teach my classmates.We will get higher marks i am sure .Thanks again .
     
  13. Dec 18, 2011 #12

    ehild

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    You are welcome. Hoping to meet you here soon again.:smile:

    ehild
     
  14. Dec 20, 2011 #13
    Hi again.My teacher told me everythink is true except elasticity coefficient ce of reduced spring - and Vo for t=0, if the law of motion x(t) is given x(t)=0,4 cos 4t+0,5 sin 4t ...Thanks again.
     
  15. Dec 20, 2011 #14

    ehild

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    If x(t) = 0.4 cos (4t)+0.5 sin(4t) the velocity is -0.4*4 sin(4t) +0.5*4 cos(4t).
    At t=0, v(0) = 2.

    ehild
     
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