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Cambridge International AS & A Level · 17 · structured

Connect SHM equations, graphs and energy

Connect SHM equations, graphs and energy · Original GioPhysics question with a detailed, mark-by-mark answer guide.

Demand
demanding
Marks
13
Topics
4
Answer
Complete
structured13 marks

An undamped oscillator of mass 0.200 kg has amplitude 0.0400 m and angular frequency 5π rad s⁻¹. At t=0 it passes through its equilibrium position x=0 in the positive direction. Time t is in seconds.

  1. (a)

    Recall a suitable displacement–time solution and the relation between acceleration and displacement in SHM. Calculate x and a at t=0.0500 s.

    4 marks
  2. (b)

    Sketch x–t, v–t and a–t over one complete cycle on a shared time scale. Mark initial values, positive/negative peaks, quarter-cycle times and explain the phase relationships.

    5 marks
  3. (c)

    Recall and calculate the total energy, then describe its transfer between kinetic and potential forms over the cycle. Give both forms when |x|=x₀/2.

    4 marks
Ready to self-mark?Reveal the detailed answer guide
Answer overview(a) x=0.0400 sin(5πt) m and a=−ω²x; at the stated time x=0.0283 m and a=−6.98 m s⁻². (b) T=0.400 s. At t=0,0.100,0.200,0.300,0.400 s: x=0,+0.0400,0,−0.0400,0; v=+0.628,0,−0.628,0,+0.628; a=0,−9.87,0,+9.87,0. (c) E=½mω²x₀²=0.0395 J; at |x|=x₀/2, E_p=0.00987 J and E_k=0.0296 J.
01

(a)

4 marks

Recall a suitable displacement–time solution and the relation between acceleration and displacement in SHM. Calculate x and a at t=0.0500 s.

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

    The initial phase selects x=x₀sin(ωt), so x=0.0400 sin(5πt) m.

  2. 2

    Recall a=−ω²x, with ω=5π rad s⁻¹.

  3. 3

    At t=0.0500 s, phase=π/4, so x=0.0400 sin(π/4)=0.0283 m.

  4. 4

    a=−(5π)²×0.028284=−6.98 m s⁻².

02

(b)

5 marks

Sketch x–t, v–t and a–t over one complete cycle on a shared time scale. Mark initial values, positive/negative peaks, quarter-cycle times and explain the phase relationships.

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

    T=2π/(5π)=0.400 s; use sinusoidal curves through the five stated phase points.

  2. 2

    x starts at zero rising, peaks positively at 0.100 s and negatively at 0.300 s.

  3. 3

    v has amplitude ωx₀=0.628 m s⁻¹ and leads x by a quarter cycle; it is zero at displacement extremes.

  4. 4

    a has amplitude ω²x₀=9.87 m s⁻² and is opposite in phase to x.

  5. 5

    Include quantities and SI units on each vertical axis; explain a is directed towards equilibrium and speed is greatest at x=0.

03

(c)

4 marks

Recall and calculate the total energy, then describe its transfer between kinetic and potential forms over the cycle. Give both forms when |x|=x₀/2.

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

    E=½mω²x₀²=0.5×0.200×(5π)²×0.0400²=0.0395 J.

  2. 2

    At equilibrium kinetic energy is maximum and potential energy relative to equilibrium is zero; at extremes all energy is potential.

  3. 3

    E_p/E=x²/x₀², so at half amplitude E_p=E/4=0.00987 J.

  4. 4

    E_k=3E/4=0.0296 J; their sum remains constant without damping.

Complete marking guidance

Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.

  1. (a) The initial phase selects x=x₀sin(ωt), so x=0.0400 sin(5πt) m.
  2. (a) Recall a=−ω²x, with ω=5π rad s⁻¹.
  3. (a) At t=0.0500 s, phase=π/4, so x=0.0400 sin(π/4)=0.0283 m.
  4. (a) a=−(5π)²×0.028284=−6.98 m s⁻².
  5. (b) T=2π/(5π)=0.400 s; use sinusoidal curves through the five stated phase points.
  6. (b) x starts at zero rising, peaks positively at 0.100 s and negatively at 0.300 s.
  7. (b) v has amplitude ωx₀=0.628 m s⁻¹ and leads x by a quarter cycle; it is zero at displacement extremes.
  8. (b) a has amplitude ω²x₀=9.87 m s⁻² and is opposite in phase to x.
  9. (b) Include quantities and SI units on each vertical axis; explain a is directed towards equilibrium and speed is greatest at x=0.
  10. (c) E=½mω²x₀²=0.5×0.200×(5π)²×0.0400²=0.0395 J.
  11. (c) At equilibrium kinetic energy is maximum and potential energy relative to equilibrium is zero; at extremes all energy is potential.
  12. (c) E_p/E=x²/x₀², so at half amplitude E_p=E/4=0.00987 J.
  13. (c) E_k=3E/4=0.0296 J; their sum remains constant without damping.

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