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SHM Question

  1. Feb 21, 2010 #1
    1. The problem statement, all variables and given/known data
    A particle of mass 0.5kg is attached to one end of a light elastic string, natural length 1.2m, it has a modulus of elasticity of 29.4N. When at rest it lies 1.4m directly beneath A. The particle is then displaced 1.75m directly below A & released from rest.

    Find the period of the simple harmonic motion which the particle moves with while the string is taut.
    Calculate the speed of P at the first instant the string becomes slack.

    2. Relevant equations
    SHM equations?


    3. The attempt at a solution

    Okay well I'm having troubles understanding this question;

    For the first half, I just plugged the numbers into the equation;

    [tex]\omega^{2} mx = \frac{\lambda x}{l}[/tex]

    which, solving for T gives me 0.897s, the correct answer. What I fail to see is the relevance of this at all? This value for the time period is completely useless in this situation as it is no where near the time period for this particular system, where the string will become slack and fall freely under gravity for over half of the period....?

    For the second half, similarly if I use the SHM equation for speed;

    [tex] v^2 = \omega^{2} (A^2 - x^2) [/tex] using A = 0.35m and x = 0.2m (above the eq'm position)

    Solving for v I get 2.01m/s, again correct. However I then decided to use energy considerations;

    Intuitively, the kinetic energy of the particle when the string becomes slack is the elastic potential energy lost minus the gravitational potential energy gained.

    [tex] K.E = E.P.E - G.P.E [/tex]

    [tex] \frac{v^2}{4} = \frac{29.4*0.55^2}{1.2} - (0.5*9.8*0.55) [/tex]

    however this yields a completely different answer for velocity, namely 4.34m/s.

    I can't see where i'm going wrong, I guess i've just got a problem imagining SHM being used for string in general, as i'm perfectly fine with spring related questions...

    Thanks in advance
     
  2. jcsd
  3. Feb 21, 2010 #2
    Okay never mind about the second bit....I forgot it was / 2l =P.

    BUT i would still love some clearance on the first bit.
     
  4. Feb 21, 2010 #3
    Could please show what you did because I am getting 1.95m/s by the same energy considerations. Elastic P.E=3.705625 J and G.P.E=2.75 J.
     
  5. Feb 21, 2010 #4
    The first bit is a tricky one. There is SHM only upto the point the string is taut so all talk of SHM would naturally end there. The particle's motion can not be described by SHM equations unless the string becomes taut again.
    Another thing- moduli of elaticity is N/m^2. What is it doing there anyway?
     
  6. Feb 21, 2010 #5
    You've got the GPE wrong 0.o no idea how..

    I'm using the equation [tex] \frac{\lambda x}{l} [/tex] as opposed to [tex] kx [/tex]

    So the unit of elasticity is N, and l is m. Are you thinking about youngs modulus?

    And yeah, it's annoying because i'm confused as it is and it doesn't really help when the questions seem to be confusing me more in terms of their wording and what they're asking you to do :/
     
  7. Feb 21, 2010 #6
    Sorry I was thinking about Young's modulus.
    But you raise a point about G.P.E. If I am right there is no absolute measure of it. My datum is at the lowest point the mass goes so when the string becomes taut
    GPE= m*g*h
    =0.5*9.8*0.55 (A+s)
     
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