The familiar mass spring system (S)

  1. A 2 kg mass is hung from the bottom of a vertical spring, causing the spring to stretch 20 cm. What is the spring constant? (Take g=10 ms-2.)
  2. This spring is now placed horizontally on a frictionless table. One end of it is held fixed and the other end is attached to a 0.5 kg mass. The mass is then moved, stretching the spring by 10 cm and then released from rest. Establish that the system obeys the SHM equation, deduce the period of oscillation, and write down the equation describing the subsequent displacement as a function of time.
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Hint

This is just what we worked through explicitly in section S6.4. Try to do it without looking back at your notes.
 

Solution

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Solution

  1. Let M represent the mass that is hung from the bottom of the spring. In equilibrium (when the mass hangs, at rest) Newton’s 2nd Law (applied to the mass M) tells us that:

    0=Fgrav-Fspring=Mg-kxe

    where xe is the equilibrium extension of the spring. You need care with signs here. Mg is the downward gravitational force; kxe is the upward force due to the spring. It follows that

    \[ k= \frac{Mg}{x_e} = \frac{2 \times 10}{0.2} =100 \,Nm^{-1} \]

  2. Now think of the familar mass-spring arrangement. And rehearse the argument for yourself once more, so you will be ready to apply it in a different context in succeding problems.
    • Identify the situation of stable equilibrium: the spring is unstretched –because the spring PE is lowest there!
    • Think, qualitatively, about the NSE behaviour: the mass will oscillate back and forth
    • Choose the coordinate x: a displacement of the mass to right, measuring spring extension
    • Choose a system to which to apply Newton’s 2nd Law: the mass, m
    Graphic - No title - shmaspring
    • Identify the force(s) acting on this system:

      F=-kx

    • Write down the equation of motion:
      \[ m\ddot{x} = -k x \]
    • Rearrange this equation in the standard form and identify ω:
      \[ \ddot{x} = -\omega^2 x \hspace{0.2cm}\mbox{\rm with}\hspace{0.2cm} \omega= \sqrt{\frac{k}{m}} \]
    • And now (but not before) you can put in numbers to give:
      \[ \omega = \sqrt{\frac{100}{0.5}} = 10\sqrt{2}\,rad\cdot s^{-1}. \]
      The period follows as
      \[ T = \frac{2\pi}{\omega} = 0.44\,s \]
    • The initial conditions are x(t=0)=0.1 and v(t=0)=0. The equation describing the subsequent displacement is then
      x=0.1cos(14.1t)
      where x is in metres and t is in seconds.