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Wave behaviour · Question 10

Wave behaviour · Original GioPhysics question with a detailed, mark-by-mark answer guide.

Demand
discriminating
Marks
14
Topics
2
Answer
Complete
Axes of energy against displacement, for the sketch−8.0−4.004.08.00displacement x / cmenergy / J
Figure 6A pair of empty axes provided for the sketch. The horizontal axis is labelled 'displacement x / cm' and is scaled from −8.0 through 0 to +8.0, with faint gridlines at −8.0, −4.0, 0, +4.0 and +8.0 and a dashed vertical line drawn at x = 0. The vertical axis is labelled 'energy / J' and carries only a zero at its foot, so no numerical scale is imposed. No curve of any kind is drawn on the axes.
structured14 marks

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.

    4 marks
  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.

    3 marks
  3. (c)

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

    4 marks
  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.

    3 marks
Ready to self-mark?Reveal the detailed answer guide
Answer overviewKey answer: ω = 2π/T = 2π/1.6 = 3.93 rad s⁻¹ For simple harmonic motion, the total energy is E_total = ½mω²x₀². E = ½ × 0.25 × 3.93² × 0.080²
01

(a)

4 marks

Determine Determine the total energy of the oscillating particle.

How to approach it

List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.

  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

02

(b)

3 marks

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.

How to approach it

List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.

  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

03

(c)

4 marks

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

How to approach it

List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.

  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

04

(d)

3 marks

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.

How to approach it

State the outcome first, then link cause to effect with the relevant physical principle. Each link in the reasoning should be explicit enough to earn its own marking point.

  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

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