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Find the Phase of a horizontal spring

  1. Dec 1, 2007 #1
    1. The problem statement, all variables and given/known data
    An air track glider attached to a spring oscillates with a period of 1.5s AT t= 0s the glider is 5.0 cm left of the equilibrium position and moving to the right at 36.3 cm/s

    so from what the question i have determined that T = 1.5s x= -0.05m V_x = 0.363m/s
    t = 0

    Find the phase constant

    2. Relevant equations

    using these numbers i sub them into the equations for x(t) = Acos( [tex]\omega[/tex]t + [tex]\varphi[/tex]) where [tex]\varphi[/tex] is the phase. -[tex]\omega[/tex]Asin( [tex]\omega[/tex]t + [tex]\varphi[/tex]) where [tex]\varphi[/tex] is the phase. and for
    [tex]\omega[/tex] i sub in 2pi / T

    3. The attempt at a solution

    so after i sub them i solve for [tex]\varphi[/tex], in each equation then i make them equal each other in an attempt to solve for A. however then i dont know what to do. i end up with a mess of inverse sin and cosines ... Am i going about this the right way? if not can some one nudge me in the right direction?
    Last edited: Dec 2, 2007
  2. jcsd
  3. Dec 1, 2007 #2
    You're going about it exactly the right way. You've just made a mistake differentiating it. I won't say how... I'll just observe that the problem would be a lot easier if that t wasn't there. And that it's an easy problem :biggrin:
  4. Dec 1, 2007 #3
    well technically the t isnt there since t = 0 wt = 0 therefore the new equation becomes

    Acos([tex]\varphi[/tex]) = x(t) and V_x(t) =-[tex]\omega[/tex]A([tex]\varphi[/tex]) and i didnt differentiate the x(t) formula into the v(t) one they were given by the textbook
    Last edited: Dec 1, 2007
  5. Dec 1, 2007 #4
    In your OP you have a t coefficient of the sine term that shouldn't be there.
    if x(t)=Acos(wt+phi)
    then v(t)=-wAsin(wt+phi)
    If the t is in front of the sine in the textbook it's a typo. Otherwise your initial velocity would be zero :biggrin:
  6. Dec 1, 2007 #5
    oh.. srry my bad i typed it wrong ill edit it now.. but when i solve for phi in each equation then make them equal i get like

    sin^-1[[tex]\frac{V_x}{A\omega}[/tex]] = cos^-1[[tex]\frac{x}{A}[/tex]]

    I cant get A out from there like if i make one side [[tex]\frac{V_x}{A\omega}[/tex]] or [[tex]\frac{x}{A}[/tex]] the other side becomes the sin or cos of the inverse cos or sin. A is still stuck in the equation. how do i rearrange to solve for A?
  7. Dec 1, 2007 #6
    You're making life really difficult for yourself like that. Look at the two equations you have. Can you see a way to eliminate the A term without solving either equation?
  8. Dec 1, 2007 #7
    i could add pi/2 to the sin angle and then change the sin^-1 to cos^1 then solve for A
  9. Dec 1, 2007 #8
    There's a much neater way of doing it. You don't have to solve any equations when you get rid of (hint:cancel) A. Have a think while I get some sleep :biggrin:
  10. Dec 1, 2007 #9
    well i just realized that A = x/cos(phi) therefore when you sub that into velocity you can get sin(phi) / cos(phi) which = tan(phi) therefore you can solve for phi however the answer that i get is only 1/2 the answer in the textbook which is -2pi/3.
  11. Dec 2, 2007 #10
    Again, you've made a slight error. But you're 95% of the way there.
    It may help to think about it in terms of dividing one equation by another, rather than making substitutions using numerical values.
    Also, I've noticed an inconsistency in your units in the OP- can you check those? And I don't get that anwer either :confused:
  12. Dec 2, 2007 #11
    The answer is in radians 360 = 2pi the negative would be that it is in the clockwise direction. but how can i get rid of A by dividing they are different equations
    isolating A then making the equations equal results in the same answer.
    the answer i got was phi = 1.04748404 rad which is ~ -pi/3
  13. Dec 2, 2007 #12
    If you know that some equation =A and some other equation=B, then you are at perfect liberty to state that (some equation)/(some other equation)=A/B. Also you made a slight mistake in your working- there should be an omega there! But even with that taken into account I'm struggling to get that textbook answer atm...
  14. Dec 2, 2007 #13
    Hmm. If your 1.047... term was positive, then that's easy: that's plus pi/3, and tan has period pi. The thing that confuses me is that it looks like there should still be an omega, which gives me the wrong answer!
  15. Dec 2, 2007 #14
    hmm.. maybe the text book is wrong? what answer are you getting? is it the same as mine?
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