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AP · AP C:M · free response

AP Physics C: Mechanics · Question 8

AP Physics C: Mechanics · Original GioPhysics question with a detailed, mark-by-mark answer guide.

Demand
discriminating
Marks
12
Topics
2
Answer
Complete
free response12 marks

A particle of mass m moves along the x-axis in a conservative field described by the potential energy function U(x) = ax⁴ − bx², where a and b are positive constants.

  1. (a)

    Derive Derive an expression for the force on the particle as a function of x.

    2 marks
  2. (b)

    Determine Determine the positions of all equilibrium points, and state for each whether it is stable or unstable.

    4 marks
  3. (c)

    Sketch Sketch a graph of U(x) against x, marking the equilibrium positions, and on the same axes indicate a total energy E for which the particle is confined to a region on one side of the origin only.

    3 marks
  4. (d)

    Derive Derive an expression for the angular frequency of small oscillations of the particle about one of the stable equilibrium positions.

    3 marks
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Answer overviewKey answer: 1 point: uses F = −dU/dx 1 point: F = −4ax³ + 2bx 1 point: equilibrium requires F = 0, so 2bx = 4ax³
01

(a)

2 marks

Derive Derive an expression for the force on the particle as a function of x.

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

    1 point: uses F = −dU/dx

  2. 2

    1 point: F = −4ax³ + 2bx

02

(b)

4 marks

Determine Determine the positions of all equilibrium points, and state for each whether it is stable or unstable.

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

    1 point: equilibrium requires F = 0, so 2bx = 4ax³

  2. 2

    1 point: x = 0 and x = ±√(b/2a)

  3. 3

    1 point: x = 0 is unstable — U has a local maximum there, since d²U/dx² = −2b < 0

  4. 4

    1 point: x = ±√(b/2a) are stable, since d²U/dx² = 12ax² − 2b = 4b > 0 at those points

03

(c)

3 marks

Sketch Sketch a graph of U(x) against x, marking the equilibrium positions, and on the same axes indicate a total energy E for which the particle is confined to a region on one side of the origin only.

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

    1 point: symmetric double-well curve, with U → +∞ as x → ±∞ and a local maximum of U = 0 at x = 0

  2. 2

    1 point: minima marked at x = ±√(b/2a), with U negative there

  3. 3

    1 point: a horizontal line drawn at a negative energy E, between the minimum value of U and zero, with the two turning points on one side identified

04

(d)

3 marks

Derive Derive an expression for the angular frequency of small oscillations of the particle about one of the stable equilibrium positions.

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

    1 point: for small displacements about a minimum, the effective spring constant is k_eff = d²U/dx² evaluated at that point

  2. 2

    1 point: k_eff = 12a(b/2a) − 2b = 4b

  3. 3

    1 point: ω = √(k_eff/m) = 2√(b/m)

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