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
Reveal
Hide
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
Reveal
Hide
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 as
E=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!
![\[ U(x_m/2) = \frac{1}{2}(65)(0.11/2)^2 = 0.098 J = E/4. \]](mastermathpng-1.png)
- The kinetic energy at this point follows from energy conservation
![\[ K = E - U =E-\frac{E}{4} = \frac{3E}{4}= 0.294\,J\hspace{0.5cm}\mbox{\rm (3sf)}\hspace{0.5cm} \]](mastermathpng-2.png)
Now let us think about the average values of K and U over the whole cycle.
The PE at any time t is
![\[ U(t)= \frac{1}{2} kx(t)^2 =\frac{1}{2} kx_m^2 \cos^2 (\omega t) =E\cos^2 (\omega t) \]](mastermathpng-3.png)
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 that
K(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 use![\[ \overline{\sin^2 (\omega t)} = \frac{1}{T} \int _0^{T} dt \sin^2 (\omega t) \]](mastermathpng-5.png)
to 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 write![\[ \overline{\sin^2 (\omega t)} = \frac{1}{2\pi}\int _0^{2\pi} dz \sin^2 (z) \]](mastermathpng-7.png)
![\[ \overline{\sin^2 (\omega t)}= \frac{1}{2\pi}\int _0^{2\pi}dz(1-cos (2z))/2 \]](mastermathpng-8.png)
The 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...?![\[ \overline{\sin^2 (\omega t)}= \frac{1}{2\pi}\int _0^{2\pi}dz \times \frac{1}{2} = \frac{1}{2} \]](mastermathpng-9.png)