Q7.8 Horizontal Mass Spring System

A block is connected to a light spring for which the force constant is 5Nm-1. The other end of the spring is fixed. The block free to oscillate on a horizontal frictionless surface. The block is displaced 5cm from equilibrium and released with an oscillation frequency of 5 rad s-1
  1. Establish that the system obeys the SHM equation.
  2. Calculate the mass of the block.
  3. Write down an equation describing the position of the block as a function of time.
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Hint

Think of the standard SHM model you have been shown in lectures and use the same principles and equations.
 

Solution

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Solution

  1. 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:
      \[ \ddot{x} = -\omega^2 x \hspace{0.2cm}\mbox{\rm with}\hspace{0.2cm} \omega= \sqrt{\frac{k}{m}} \]
  2. Combining the above we can show that

    \[ m\omega^2x=kx \hspace{0.5cm}\mbox{\rm which can be rearranged to give}\hspace{0.5cm} m = \frac{k}{\omega^2} \]

    Putting in the numbers

    \[ m = \frac{5}{(5)^2} = 0.2kg\]

  3. The initial conditions are x(t=0)=0.05 and v(t=0)=0. The equation describing the subsequent displacement is then
    x=0.05cos(5t)
    where x is in metres and t is in seconds.