OCR · A-Level · Physics

    Course version: H556 · Version from Sept 2015

    (d) solutions to the equation a = -ω²x e.g. x = A cos(ωt) or x = A sin(ωt)

    Full specification wording

    Simple harmonic oscillations: (d) solutions to the equation a = -ω²x e.g. x = A cos(ωt) or x = A sin(ωt)

    The defining equation of simple harmonic motion is a = -ω²x, where a is acceleration, x is displacement from equilibrium and ω is angular frequency.

    Checked against the OCR H556 specification (opens in a new tab) on .

    • Point 4 of 7
    • 4 key terms

    Study notes

    From the checked teaching notes for this specification point.

    The idea

    The defining equation of simple harmonic motion is a = -ω²x, where a is acceleration, x is displacement from equilibrium and ω is angular frequency. Its solutions are sinusoidal functions of time: x = A cos(ωt) or x = A sin(ωt), where A is the amplitude. Differentiating x = A cos(ωt) gives v = -Aω sin(ωt) and a = -Aω² cos(ωt), which equals -ω²x. The cosine form starts at maximum displacement when t = 0; the sine form starts at equilibrium moving in the positive direction. Both describe the same motion with a different choice of starting time, and the period is T = 2π/ω.

    What earns the marks

    • Recognises that x = A cos(ωt) and x = A sin(ωt) are both solutions of a = -ω²x.
    • Identifies A as the amplitude and ω as the angular frequency, with ω = 2πf = 2π/T.
    • Differentiates a displacement solution to obtain velocity and acceleration, showing that a = -ω²x follows.
    • Distinguishes the initial conditions: cosine starts at maximum displacement, sine starts at equilibrium.
    • Uses the relationship between ω, period and frequency to convert between forms.

    Common mistakes and how to put them right

    • Treating ω as the frequency f rather than the angular frequency.

      Put it right: Use ω = 2πf.

    • Believing that only one of the sine and cosine forms is a valid solution.

      Put it right: Recognise that both satisfy the equation and differ only in initial conditions.

    • Dropping the minus sign in a = -ω²x, which reverses the direction of the restoring acceleration.

      Put it right: Keep the negative sign because acceleration is always directed towards equilibrium.

    • Confusing amplitude A with displacement x.

      Put it right: Note that x varies with time while A is the maximum value of x.

    Exam tips

    • Check the initial condition described in the question before choosing sine or cosine.
    • Convert between period and angular frequency using ω = 2π/T before substituting into a solution.
    • Verify a proposed solution by differentiating twice and comparing with a = -ω²x.

    Try the lesson

    The start of the Simple harmonic oscillations: (d) solutions to the equation a = -ω²x e.g. x = A cos(ωt) or x = A sin(ωt) lesson, made from checked teaching notes for this specification point.

    Listen

    The defining equation of simple harmonic motion is a = -ω²x, where a is acceleration, x is displacement from equilibrium and ω is angular frequency. Its solutions are sinusoidal functions of time: x = A cos(ωt) or x = A sin(ωt), where A is the amplitude. Differentiating x = A cos(ωt) gives v = -Aω sin(ωt) and a = -Aω² cos(ωt), which equals -ω²x. The cosine form starts at maximum displacement when t = 0; the sine form starts at equilibrium moving in the positive direction. Both describe the same motion with a different choice of starting time, and the period is T = 2π/ω. A strong answer earns marks for these points. Recognises that x = A cos(ωt) and x = A sin(ωt) are both solutions of a = -ω²x. Identifies A as the amplitude and ω as the angular frequency, with ω = 2πf = 2π/T. Differentiates a displacement solution to obtain velocity and acceleration, showing that a = -ω²x follows. Distinguishes the initial conditions: cosine starts at maximum displacement, sine starts at equilibrium. A common slip: Treating ω as the frequency f rather than the angular frequency. Instead, use ω = 2πf. Exam tip: Check the initial condition described in the question before choosing sine or cosine.

    See it

    A top-down flowchart. At the top is the node a = -ω²x, which branches to two nodes: x = A cos(ωt) and x = A sin(ωt). From x = A cos(ωt) a chain runs down: to v = -Aω sin(ωt), then to a = -Aω² cos(ωt), then to a = -ω²x. Also from x = A cos(ωt) a branch leads to the node 'Cosine starts at maximum displacement'. From x = A sin(ωt) a branch leads to the node 'Sine starts at equilibrium'.
    Shows that both x = A cos(ωt) and x = A sin(ωt) solve a = -ω²x, with the cosine form differentiated twice back to a = -ω²x, and the differing initial conditions of the two forms.

    Key words

    Say what each word means before you turn the card.

    Angular frequency

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    Amplitude

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    Cosine form

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    Sine form

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    See every definition
    Angular frequency
    The quantity ω in the solutions, linked to period and frequency by ω = 2πf = 2π/T.
    Amplitude
    The quantity A in x = A cos(ωt) or x = A sin(ωt), being the maximum value of displacement x.
    Cosine form
    The solution x = A cos(ωt), which starts at maximum displacement when t = 0.
    Sine form
    The solution x = A sin(ωt), which starts at equilibrium moving in the positive direction.

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