Q7.15 Energy in SHM (S,H)
A mass-spring system comprises a block of mass 0.68 kg, fastened to a spring of spring constant 65 Nm-1. The block is pulled a distance x=xm=0.11 m from equilibrium and released.- Determine the total energy of the oscillator.
- Determine the potential energy when x=0.5xm.
- Determine the kinetic energy when x=0.5xm.
- What can you say about the average kinetic and potential energies, over a complete cycle?
Hint
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Hint
Remind yourself of Key Point 6.7. In the last part you may feel the need to work out the ‘average’ of the square of a sinusoidal function, over one period. You can do this by integration. Or you can avoid the maths by a bit of extra thought. Choose.Solution
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Solution
- The total
energy E is the sum of potential energy U and kinetic energy K.
E=U+KThe block is released from rest. At this point K=0. Its initial displacement will define the amplitude of the motion; the PE at this point isThe total energy follows asE=U(xm)=0.393 Jand remains constant throughout the motion (Key Point 6.7).
- Setting x=xm/2 we find
Note that halving the displacement does not halve the PE!
- The kinetic energy at this point follows from energy conservation
Now let us think about the average values of K and U over the whole cycle.
The PE at any time t is
The KE at any time t is
where the last step uses ω2=k/m (one of those results you should be carrying around in your head, by now!) You get confident that you are on the right track by noting thatK(t)+U(t)=E[cos2(ωt)+sin2(ωt)]=EIn each case then the average we want is of the form:
E×average of square of a sinusoidal functionIn the one case it is a sine function, in the other a cosine; but it ought to be clear that the averages will be the same: after all a sine function is just a shifted cosine function. The only way in which the averages can be the same and the total have constant value E is if each has average value E/2.If you prefer a mathematically tight argument to a wordy one, then here it is:
The average of (say) sin2(ωt) over one cycle is:
where we useto mean ‘average’ of ‘thing’. The right hand side effectively adds up the value of the sin2 at a whole series of equally spaced intervals and then divides by the number of the intervals. Now, introduce the new variable z=2πt/T. Then
Using sin2z=(1-cos(2z))/2 (which some of you do, or one day will, carry around in your head) we can writeThe integral over a cosine (as distinct from the square of a cosine) over one cycle must give zero, so we are left with
which recovers the result we argued our way to above. Perhaps you preferred the argument without the mathematics...?