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- Establish that the system obeys the SHM equation.
- Calculate the mass of the block.
- Write down an equation describing the position of the block as a function of time.
Hint
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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
- 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
Identify the force(s) acting on this system:
F=-kx- Write down the equation of motion:
- Rearrange this equation in the standard form and identify ω:
Combining the above we can show that
Putting in the numbers
- 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.