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Getting different result with energy conservationusing and Newton's 2nd law(in SHM)

  1. Oct 28, 2012 #1
    Hello guys i was trying to solve some problem of simple harmonic motion.
    as it is well explaind in the title i'm not getting the same result as my textbook manual solution.

    Problem:
    A block of unknown mass is attached to a spring with a spring constant of 6:50 N/m and undergoes simple harmonic motion with an amplitude of 10:0 cm. When the block is halfway between its equilibrium position and the end point, its speed is measured to be 30:0 cm/s. Calculate (a) the mass of the block, (b) the period of the motion, and (c) the maximum acceleration of the block.

    Textbook solution:
    (a) Conserving energy
    E =1/2kA² = 32:5 mJ
    E =1/2k(A/2)²+12mv2 = E/4+1/2mv²
    m =(2/v²)(3/4)E =3E/(2v²) = 542 g

    My attempt with SHM equation :
    x(t) = Asin(ωt + ∅) and ω=[itex]\sqrt{}(k/m)[/itex]
    x(t) = Asin(ωt + ∅)= A/2
    sin(ωt + ∅) = 1/2 → ωt + ∅ = ∏/6 → t = (∏/6 - ∅)/ω

    We have as well: v(t)=dx/dt= Aωcos(ωt + ∅)
    And at x=A/2 → v(t)=0.3 m/s
    Substituting t in v(t) we'll have Aωcos(∏/6) = 0.3 → after computation done and substituting the value of A and k i find another number different from 542 g.

    Am I wrong? Surely yes, but where? Thanks in advance for helping me :)
     
  2. jcsd
  3. Oct 28, 2012 #2

    Simon Bridge

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    Re: Getting different result with energy conservationusing and Newton's 2nd law(in SH

    I take it you are using a colon for a decimal point?
    Check your working - looks like you've substituted something wrong or made an algebraic error where you havn't shown your working.
     
  4. Oct 28, 2012 #3

    Simon Bridge

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    Re: Getting different result with energy conservationusing and Newton's 2nd law(in SH

    Your final step was from ##v=\omega A\cos(\omega t)##, which gives you $$v=\frac{\sqrt{3}}{2}\sqrt{\frac{k}{m}} A$$... because you found that ##\omega t=\frac{\pi}{6}## and ##\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}##

    The task is to find m given the rest.

    square both sides and solve for m gives:$$m=\frac{3}{4}\frac{kA^2}{v^2}$$

    CoE starts out with:
    $$\frac{1}{2}mv^2=\frac{1}{2}kA^2 - \frac{1}{2}k\left (\frac{A}{2}\right )^2$$... (do you understand how this equation happens?) cancel the halves and group common terms: $$mv^2=\frac{3}{4}kA^2$$... same equation.

    Work through them slowly by hand - step-by-step.
     
  5. Oct 29, 2012 #4
    Re: Getting different result with energy conservationusing and Newton's 2nd law(in SH

    Thanls a lot, it was just a calculation error.
     
  6. Oct 29, 2012 #5

    Simon Bridge

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    Re: Getting different result with energy conservationusing and Newton's 2nd law(in SH

    No worries - it can be hard to spot these yourself.
    The best way to avoid them (or, at least, pick them up as they happen) is to do all the algebra at the start - when you have the final equation, then you substitute in the values.
    You'll start noticing more experienced people doing just that - even when putting the numbers in sooner can make the calculation seem easier.
     
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