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Particle moving with simple harmonic motion

  1. Sep 19, 2016 #1
    1. The problem statement, all variables and given/known data
    A particle moves with simple harmonic motion in a straight line with amplitude 0.05 m and period 12 s. Find: (a) the maximum speed, (b) the maximum acceleration, of the particle.

    Write down the values of the constants P and Q in the equation x / m = P sin [Q (t / s)] which describes its motion.

    Answers: (a) 0.026 m s-1, (b) 0.014 m s-2, P = 0.05, Q = π / 6

    2. The attempt at a solution
    (a) I used the v = (2 π r) / T formula for this part: v = (2 * π * 0.05) / 12 = 0.026 m s-1.

    (b) For acceleration I used this formula: a = v2 / r → a = 0.0262 / 0.05 = 0.0137 m s-2.

    In terms of P and Q I am somewhat lost. Let's take a look at the formula provided:
    • I think x is the extension = 0.05 m
    • m is the mass, which is unknown and I don't know how to find it (I can only think of KE, but in that case I'll have two unknowns -- KE and m)
    • Both P and Q are unknown, I think we need to plug everything in and then derive P (P = (x / m) / sin [Q (t / s)]) and then plug P into the original equation, but not sure
    • t should be time = 12 s
    • s should be the distance moved s = v t = 0.026 * 0.05 = 1.3 * 10-3 m

    In sum, if my logic is correct, I don't see a way how to find m. And I my logic is wrong, how should I approach the given formula?
     
  2. jcsd
  3. Sep 19, 2016 #2

    Doc Al

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    Staff: Mentor

    I think m and s just represent "meters" and "seconds". m is not mass! To me, that's a confusing way of describing things, but they want P and Q to be without units.

    Think of the equation as: x = P sin [Q t], then it should make sense. (Compare that with the standard form of the position as a function of time for SHM.) All the info needed to find P and Q is in the problem statement.
     
  4. Sep 19, 2016 #3

    Ray Vickson

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    I think the notation in the question is horrible. Perhaps ##m## stands for "meter" and ##s## for second, but why they would want to write it like that is a mystery to me. A better way would be to say that ##x(t) = a \sin(\omega t)## and say that ##x, a## have units of "meters", ##t## has units of s = seconds, while ##\omega## has units of 1/s = per second. Maybe ##P s = a## and ##Q/s = \omega##.

    Note added in edit: Doc Al's post #2 above did not appear on my screen until after I pressed the enter key to submit this message.
     
  5. Sep 19, 2016 #4
    Hm, if we simplify the formula to x = P sin (Q t) it in fact makes more sense. We have x = 0.05 m and t = 12 s. P = 0.05 / sin (12 Q). Plug it into the original equation: 0.05 = (0.05 / sin (12 Q)) * sin (12 Q)
    0.05 sin (12 Q) = 0.05 sin (12 Q)

    Am I missing something?

    Update: actually the problem is very weird. If I am required to find ω, then it's just ω = 2 π / T and I get the required answer π / 6. And it looks like P is the amplitude, which is already given and is equal to 0.05 m.
     
    Last edited: Sep 19, 2016
  6. Sep 19, 2016 #5

    Doc Al

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    The "standard form" that I mentioned is the equation that Ray Vickson gave you: ## x(t) = a \sin(\omega t)##. Note that "a" is P and ω is Q.

    Hint: a is given directly, and ω should be easy to figure out.
     
  7. Sep 19, 2016 #6
    I am even more confused now. The original question was about transforming the formula into this one and deriving P just by looking at the given 0.05 amplitude and deriving ω by ω = 2 π / T = π / 6? Or do I need to plug everything in? x(t) will be 0.05 m?
     
  8. Sep 19, 2016 #7

    Doc Al

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    Exactly. That's all there is to it. They threw you off with that weird business with the "m" and the "s". (The purpose of that was to get rid of the units. So P = 0.05, not 0.05 m. Seems silly--and confusing--to me.)
     
  9. Sep 19, 2016 #8
    Alright, indeed a very confusing problem definition.
     
  10. Sep 19, 2016 #9

    Doc Al

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    That's right.

    Nope. Nothing that complicated.
     
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