Theme C · Wave behaviour

The amplitude of a mass–spring oscillator is doubled while its mass and spring constant are unchanged. What happens to its period and to its total energy?

1. Aperiod unchanged, energy doubled
2. Bperiod unchanged, energy quadrupled
3. Cperiod doubled, energy doubled
4. Dperiod doubled, energy quadrupled

A sound wave of frequency 256 Hz travels through air at 340 m s⁻¹ and then passes into water, where its speed is 1500 m s⁻¹. What are its frequency and wavelength in the water?

1. A256 Hz and 1.33 m
2. B256 Hz and 5.86 m
3. C1130 Hz and 1.33 m
4. D58 Hz and 5.86 m

A pipe of length 0.85 m is closed at one end and open at the other. The speed of sound in air is 340 m s⁻¹. What is the frequency of the fundamental note?

1. A100 Hz
2. B200 Hz
3. C400 Hz
4. D800 Hz

An ambulance siren emits a note of frequency 512 Hz. The ambulance travels at 25 m s⁻¹ towards a stationary observer. The speed of sound in air is 340 m s⁻¹. What frequency does the observer hear?

1. A477 Hz
2. B512 Hz
3. C553 Hz
4. D590 Hz

Monochromatic light passes through a single slit and forms a diffraction pattern on a distant screen. The slit is then made narrower. What happens to the width of the central maximum and to its intensity?

1. Awidth increases, intensity decreases
2. Bwidth increases, intensity increases
3. Cwidth decreases, intensity decreases
4. Dwidth decreases, intensity increases

A particle undergoes simple harmonic motion described by x = x₀ sin(ωt). At what displacement is the kinetic energy equal to the potential energy?

1. Ax₀/4
2. Bx₀/2
3. Cx₀/√2
4. Dx₀

A student investigates the oscillations of a mass hanging from a vertical spring, expecting the period to satisfy T = 2π√(m/k), where k is the spring constant.

1. (a)

   Determine Determine the gradient of a graph of T² against m, and hence determine the spring constant k.

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   1. 1

      T² = (4π²/k)m, so the gradient is 4π²/k

   2. 2

      gradient = (0.987 − 0.197)/(0.500 − 0.100) = 0.790/0.400

   3. 3

      gradient = 1.98 s² kg⁻¹

   4. 4

      k = 4π²/1.98 = 20 N m⁻¹

2. (b)

   Explain Explain why the student plotted T² against m rather than T against m.

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   1. 1

      T against m would be a curve, from which no constant can be read directly

   2. 2

      squaring produces a linear relationship, so a straight line both tests the model and gives k from a single gradient

3. (c)

   Suggest In practice the line of best fit for a real spring has a small positive intercept on the T² axis. Suggest a physical reason for this.

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   1. 1

      the spring itself has mass, and part of it oscillates along with the hanging mass

   2. 2

      so the effective oscillating mass exceeds the mass added, giving a non-zero period as the added mass tends to zero — the intercept corresponds to roughly one third of the spring's own mass

4. (d)

   Outline Outline how the student should time the oscillations to keep the uncertainty in T small, and state one reason why timing a single oscillation would be unsatisfactory.

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   1. 1

      time at least 10 or 20 complete oscillations and divide by the number

   2. 2

      start and stop the timing as the mass passes through the equilibrium position, where it moves fastest, using a fixed reference marker

   3. 3

      the reaction-time uncertainty of about 0.2 s is a large fraction of a single period of about 0.5 s, but only a small fraction of a 10-oscillation total

A string of length 0.60 m is stretched between two fixed points. When plucked, it vibrates in its fundamental mode at a frequency of 250 Hz.

1. (a)

   Determine Determine the speed of the transverse waves on the string.

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   1. 1

      in the fundamental mode the string carries half a wavelength, so λ = 2L = 1.2 m

   2. 2

      v = fλ = 250 × 1.2

   3. 3

      v = 300 m s⁻¹

2. (b)

   Describe Describe how a standing wave is formed on the string, and state what distinguishes a standing wave from a progressive wave.

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   1. 1

      a wave travels along the string and reflects at the fixed end, and the reflected wave travels back along the string

   2. 2

      the two waves have the same frequency, speed and amplitude and travel in opposite directions, so they superpose

   3. 3

      the superposition produces fixed nodes where the two always cancel and antinodes where they always reinforce

   4. 4

      a standing wave transfers no energy along the string, and every point between adjacent nodes oscillates in phase — unlike a progressive wave, in which phase changes steadily along the direction of travel

3. (c)

   Calculate Calculate the frequency of the third harmonic of this string, and state how many nodes it has, including the two at the ends.

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   1. 1

      f₃ = 3 × 250 = 750 Hz

   2. 2

      The third harmonic has four displacement nodes, including both fixed ends.

4. (d)

   Explain The tension in the string is increased. Explain what happens to the fundamental frequency, and explain why the length of the string does not change the speed of the waves on it.

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   1. 1

      the wave speed on a string increases with tension, so v rises

   2. 2

      the wavelength of the fundamental is fixed at 2L by the geometry, so f = v/λ increases

   3. 3

      the speed depends only on the tension and the mass per unit length of the string — properties of the medium — and not on how much of that medium is between the supports

Coherent light of wavelength 633 nm from a laser is incident normally on a pair of narrow parallel slits separated by 0.25 mm. An interference pattern is observed on a screen 2.40 m from the slits.

1. (a)

   Determine Determine the separation of adjacent bright fringes on the screen.

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   1. 1

      For small-angle double-slit interference, adjacent fringe separation is s = λD/d.

   2. 2

      s = (633 × 10⁻⁹ × 2.40)/(0.25 × 10⁻³)

   3. 3

      s = 6.1 × 10⁻³ m (6.1 mm)

2. (b)

   Explain Explain, in terms of path difference, why a bright fringe appears at the centre of the pattern and why dark fringes appear either side of it.

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   1. 1

      at the centre the two paths are equal, so the path difference is zero and the waves arrive in phase and interfere constructively

   2. 2

      moving away from the centre, the path difference grows

   3. 3

      where it reaches half a wavelength the waves arrive antiphase and cancel, giving a dark fringe; where it reaches a whole wavelength they reinforce again

3. (c)

   Suggest One of the two slits is now covered with a filter that halves the amplitude of the light passing through it. Suggest how the appearance of the pattern changes.

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   1. 1

      the fringe separation is unchanged, because it depends only on λ, D and d

   2. 2

      the dark fringes are no longer completely dark, because the two amplitudes no longer cancel exactly

   3. 3

      so the contrast between bright and dark fringes is reduced, and the bright fringes are dimmer

4. (d)

   Compare The double slit is replaced by a diffraction grating of 500 lines per millimetre, illuminated by the same laser. Compare the pattern produced with the double-slit pattern.

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   1. 1

      the grating produces maxima that are much sharper and narrower, separated by wide dark regions, rather than broad fringes of gradually varying brightness

   2. 2

      the maxima are much further apart, because the slit separation of 2.0 × 10⁻⁶ m is far smaller than 0.25 mm

   3. 3

      each maximum is brighter, because light from many thousands of slits contributes to it rather than from only two

A particle of mass 0.25 kg undergoes simple harmonic motion with amplitude 8.0 cm and period 1.6 s. Separately, a star in a distant galaxy emits a spectral line of laboratory wavelength 486.1 nm, which is observed on Earth at 487.5 nm.

1. (a)

   Determine Determine the total energy of the oscillating particle.

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   1. 1

      ω = 2π/T = 2π/1.6 = 3.93 rad s⁻¹

   2. 2

      For simple harmonic motion, the total energy is E_total = ½mω²x₀².

   3. 3

      E = ½ × 0.25 × 3.93² × 0.080²

   4. 4

      E = 1.2 × 10⁻² J

2. (b)

   Sketch Sketch, on the same axes, graphs of the kinetic energy and of the potential energy of the particle against displacement, from −x₀ to +x₀.

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   1. 1

      potential energy is a parabola with a minimum of zero at x = 0, rising to the total energy at x = ±x₀

   2. 2

      kinetic energy is an inverted parabola, maximum at x = 0 and zero at x = ±x₀

   3. 3

      the two curves sum to a constant at every displacement, and cross at x = ±x₀/√2

3. (c)

   Determine Determine the speed at which the star is moving relative to the Earth, and state whether it is approaching or receding.

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   1. 1

      Δλ = 487.5 − 486.1 = 1.4 nm

   2. 2

      Δλ/λ₀ = v/c, so v = c(1.4/486.1)

   3. 3

      v = 8.6 × 10⁵ m s⁻¹

   4. 4

      the observed wavelength is longer, so the light is redshifted and the star is receding

4. (d)

   Explain Explain why the Doppler shift for light is treated differently from the Doppler shift for sound when the source and the observer are both moving.

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   1. 1

      sound travels through a medium, so the equations distinguish a moving source from a moving observer — the two give different shifts for the same relative speed

   2. 2

      light needs no medium, and its speed is the same in every inertial frame

   3. 3

      so only the relative velocity of source and observer can matter, and a single expression covers every case