Oscillations for IB Physics HL

Reading cue. The hatched wall is a displacement node and the unblocked end an antinode; L = 0.85 m is the whole pipe, not the quarter wavelength it holds.

1. aState State how many displacement nodes and how many displacement antinodes the air column contains when the pipe sounds its lowest note, and say where each one lies on the drawing.

   recall2 marks

   Check answer · 2 marks

   1. one node, at the closed (hatched) end
   2. one antinode, at the open end

2. bCalculate Calculate the wavelength and the frequency of the next standing wave this air column can support above its lowest note.

   routine3 marks

   Check answer · 3 marks

   1. next mode fits three quarter-wavelengths into the pipe, so λ = 4L/3
   2. λ = 4 × 0.85 / 3 = 1.13 m
   3. f = v / λ = 3 × 340 / (4 × 0.85) = 300 Hz

3. cDetermine The open end is now sealed with a second wall, so that both ends of the 0.85 m pipe are closed. Determine the frequency of the lowest note the pipe can then sound, and state how it compares with the value for the pipe as drawn.

   demanding4 marks

   Check answer · 4 marks

   1. both ends are now displacement nodes, so the pipe holds half a wavelength: L = λ/2
   2. λ = 2 × 0.85 = 1.70 m
   3. f = 340 / 1.70 = 200 Hz
   4. this is twice the 100 Hz of the pipe as drawn

4. dSuggest When this pipe is built and blown, its lowest note is measured to be a few hertz below the value predicted from the 0.85 m marked on the drawing. Suggest a physical reason for the discrepancy.

   top of the paper3 marks

   Check answer · 3 marks

   1. the displacement antinode does not form exactly at the mouth but a little way outside it
   2. the effective length of the air column is therefore greater than the 0.85 m marked between the ticks
   3. a greater effective length gives a longer wavelength, and so a frequency below the predicted value

Reading cue. The oscillation arrow is centred on the hanging position, not the spring's natural length: x is measured from there, where the spring already carries an extension mg/k.

1. aState State the point of the oscillation drawn at which the block's acceleration is zero, and the points at which its magnitude is greatest.

   recall2 marks

   Check answer · 2 marks

   1. acceleration is zero at the hanging position, at the middle of the double-headed arrow
   2. acceleration is greatest at the two ends of the arrow, at maximum displacement

2. bShow (that) Show that, before it is set oscillating, the block stretches the spring by mg/k, and hence calculate this extension for m = 0.20 kg and k = 25 N m⁻¹.

   routine3 marks

   Check answer · 3 marks

   1. at rest the upward spring tension balances the weight: ke = mg
   2. rearranging gives e = mg / k
   3. e = 0.20 × 9.8 / 25 = 7.8 × 10⁻² m (7.8 cm)

3. cDetermine For the same block and spring, determine the period of the oscillation, and determine the magnitude of the block's acceleration at the top of the arrow when the amplitude is 3.0 cm.

   demanding4 marks

   Check answer · 4 marks

   1. ω = sqrt(k/m) = sqrt(25 / 0.20) = 11.2 rad s⁻¹
   2. T = 2π / ω = 0.56 s
   3. using a = -ω² x, the magnitude at maximum displacement is ω² x0 = 125 × 0.030
   4. a = 3.8 m s⁻², directed downward, back towards the hanging position

4. dExplain Explain why, for this apparatus, the motion stops being simple harmonic once the amplitude exceeds the extension found earlier.

   top of the paper4 marks

   Show a hint

   What force can act on the block once the spring has returned to its natural length?

   Check answer · 4 marks

   1. at an amplitude equal to the static extension e, the maximum acceleration is ω² e = (k/m)(mg/k) = g
   2. at the top of that swing the spring has returned to its natural length and exerts no force on the block
   3. a larger amplitude would demand a downward acceleration greater than g, which gravity alone cannot supply
   4. the spring goes slack instead, the restoring force is no longer proportional to displacement, and the block briefly falls freely

Reading cue. The horizontal axis is displacement, not time: kinetic energy is an inverted parabola peaking at x = 0 here, not a cosine curve, and the vertical axis carries no scale.

1. aState The axes are drawn to span the whole of the motion, from one extreme to the other. State the values of x, read from the horizontal scale, at which a potential-energy curve drawn on these axes would reach the total energy of the particle, and state the kinetic energy there.

   recall2 marks

   Check answer · 2 marks

   1. at x = -8.0 cm and x = +8.0 cm, the ends of the scale, which are the amplitude
   2. the kinetic energy is zero at those two displacements

2. bDetermine The total energy of the particle is 1.23 × 10⁻² J. Determine its potential energy and its kinetic energy at the gridline x = +4.0 cm.

   routine3 marks

   Check answer · 3 marks

   1. at that gridline x / x0 = 4.0 / 8.0 = 0.50, so x² / x0² = 0.25
   2. potential energy = 0.25 × 1.23 × 10⁻² = 3.1 × 10⁻³ J
   3. kinetic energy = 1.23 × 10⁻² - 3.1 × 10⁻³ = 9.2 × 10⁻³ J

3. cDetermine Determine the displacement, on the scale given, at which the kinetic and potential energies of the particle are equal, and state whether this falls on the +4.0 cm gridline drawn.

   demanding4 marks

   Check answer · 4 marks

   1. equal energies means each is half the total, so x² / x0² = 0.50
   2. x = x0 / sqrt(2) = 8.0 / sqrt(2)
   3. x = +/- 5.7 cm
   4. this lies outside the +/- 4.0 cm gridlines, so the curves do not cross at half the amplitude

4. dExplain The particle is restarted with amplitude 4.0 cm, the gridline on these axes, and the same period. Explain what happens to each of the two energy curves, giving the new total energy.

   top of the paper4 marks

   Check answer · 4 marks

   1. total energy is proportional to x0², so it falls to a quarter: 3.1 × 10⁻³ J
   2. the potential-energy curve keeps exactly the same shape, since potential energy depends on x and not on the amplitude; it is simply followed only out to +/- 4.0 cm
   3. the kinetic-energy curve is a new, lower inverted parabola, still peaking at x = 0 but now at 3.1 × 10⁻³ J
   4. the curves still cross where each is half the total, now at x = 4.0 / sqrt(2) = +/- 2.8 cm