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.
  1. Determine the total energy of the oscillator.
  2. Determine the potential energy when x=0.5xm.
  3. Determine the kinetic energy when x=0.5xm.
  4. What can you say about the average kinetic and potential energies, over a complete cycle?
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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

  1. The total energy E is the sum of potential energy U and kinetic energy K.
    E=U+K
    The 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 is
    \[ U= U(x_m) = \frac{1}{2}kx_m^2 = \frac{1}{2}(65)(0.11)^2 = 0.393\, J \hspace{0.5cm}\mbox{\rm (3sf)}\hspace{0.5cm} \]
    The total energy follows as
    E=U(xm)=0.393 J
    and remains constant throughout the motion (Key Point 6.7).
  2. Setting x=xm/2 we find
    \[ U(x_m/2) = \frac{1}{2}(65)(0.11/2)^2 = 0.098 J = E/4. \]
    Note that halving the displacement does not halve the PE!
  3. The kinetic energy at this point follows from energy conservation
    \[ K = E - U =E-\frac{E}{4} = \frac{3E}{4}= 0.294\,J\hspace{0.5cm}\mbox{\rm (3sf)}\hspace{0.5cm} \]
  4. Now let us think about the average values of K and U over the whole cycle.

    The PE at any time t is

    \[ U(t)= \frac{1}{2} kx(t)^2 =\frac{1}{2} kx_m^2 \cos^2 (\omega t) =E\cos^2 (\omega t) \]

    The KE at any time t is

    \[ K(t)= \frac{1}{2} m v(t)^2 = \frac{1}{2} m \dot{x}(t)^2 =\frac{1}{2} m\omega^2 x_m^2 \sin^2 (\omega t) =E\sin^2 (\omega t) \]
    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)]=E

    In each case then the average we want is of the form:

    E×average of square of a sinusoidal function
    In 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:

    \[ \overline{\sin^2 (\omega t)} = \frac{1}{T} \int _0^{T} dt \sin^2 (\omega t) \]
    where we use $\overline{\mbox{thin}}$ 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
    \[ \overline{\sin^2 (\omega t)} = \frac{1}{2\pi}\int _0^{2\pi} dz \sin^2 (z) \]
    Using sin2z=(1-cos(2z))/2 (which some of you do, or one day will, carry around in your head) we can write
    \[ \overline{\sin^2 (\omega t)}= \frac{1}{2\pi}\int _0^{2\pi}dz(1-cos (2z))/2 \]

    The integral over a cosine (as distinct from the square of a cosine) over one cycle must give zero, so we are left with

    \[ \overline{\sin^2 (\omega t)}= \frac{1}{2\pi}\int _0^{2\pi}dz \times \frac{1}{2} = \frac{1}{2} \]
    which recovers the result we argued our way to above. Perhaps you preferred the argument without the mathematics...?