AQA · A-Level · Physics
Course version: 7408 · Version from Sept 2016
Appreciation that the v − t graph is derived from the gradient of the x −t…
Full specification wording
Simple harmonic motion (SHM) (A-level only): Appreciation that the v − t graph is derived from the gradient of the x −t graph and that the a − t graph is derived from the gradient of the v − t graph.
Displacement, velocity and acceleration are linked by successive gradients.
Checked against the AQA 7408 specification (opens in a new tab) on .
- Point 2 of 2
- 3 key terms
Study notes
From the checked teaching notes for this specification point.
The idea
Displacement, velocity and acceleration are linked by successive gradients. Plotting displacement x against time t, the gradient at any instant equals the velocity v, making the v–t graph the gradient function of the x–t graph. Differentiating again, the gradient of the v–t graph equals the acceleration a. For SHM, x = A cos(ωt) gives v = −Aω sin(ωt) and a = −Aω² cos(ωt) = −ω²x. Graphically, a maximum or minimum of x occurs where v is zero; the steepest x–t slope occurs where v is maximum; and a is opposite in sign to x. This explains why acceleration always points towards equilibrium, opposing velocity only when moving away from equilibrium.
What earns the marks
- State that the gradient of a displacement–time graph at a given instant gives the instantaneous velocity.
- State that the gradient of a velocity–time graph at a given instant gives the instantaneous acceleration.
- Apply the chain of gradients to SHM: where x is maximum, gradient is zero so v = 0; where x crosses zero, gradient is steepest so v is maximum in magnitude.
- Relate the sign of the v–t gradient to the direction of acceleration, showing that a is directed towards the equilibrium position and is opposite in sign to x.
- Use the standard SHM expressions x = A cos(ωt), v = −Aω sin(ωt) and a = −Aω² cos(ωt) to confirm the graphical relationships and the result a = −ω²x.
Common mistakes and how to put them right
Assuming acceleration always opposes velocity.
Put it right: In SHM, acceleration points towards equilibrium, so it opposes velocity when moving away from equilibrium, but acts in the same direction when moving towards it.
Confusing the value of the graph with its gradient.
Put it right: Velocity is the gradient of the x–t graph, not its displacement value.
Treating the gradient as constant over a curved section.
Put it right: For SHM the gradient changes continuously, so the instantaneous gradient must be found at a point, for example by drawing a tangent.
Exam tips
- Sketch x–t, v–t and a–t on the same time axis so that maxima, minima and zero crossings line up; this makes the gradient relationships visible.
- When asked to compare graphs, quote the gradient relationship explicitly, for example state that the value of v at a time equals the gradient of x–t at that time.
Try the lesson
The start of the Simple harmonic motion (SHM) (A-level only): Appreciation that the v − t graph is derived from the gradient of the x −t graph and that the a − t graph is derived from the gradient of the v − t graph. lesson, made from checked teaching notes for this specification point.
Have a go first
A displacement–time graph for an oscillating mass is plotted. How is the velocity–time graph obtained from it?
Listen
Displacement, velocity and acceleration are linked by successive gradients. Plotting displacement x against time t, the gradient at any instant equals the velocity v, making the v–t graph the gradient function of the x–t graph. Differentiating again, the gradient of the v–t graph equals the acceleration a. For SHM, x = A cos(ωt) gives v = −Aω sin(ωt) and a = −Aω² cos(ωt) = −ω²x. Graphically, a maximum or minimum of x occurs where v is zero; the steepest x–t slope occurs where v is maximum; and a is opposite in sign to x. This explains why acceleration always points towards equilibrium, opposing velocity only when moving away from equilibrium. A strong answer earns marks for these points. State that the gradient of a displacement–time graph at a given instant gives the instantaneous velocity. State that the gradient of a velocity–time graph at a given instant gives the instantaneous acceleration. Apply the chain of gradients to SHM: where x is maximum, gradient is zero so v = 0; where x crosses zero, gradient is steepest so v is maximum in magnitude. Relate the sign of the v–t gradient to the direction of acceleration, showing that a is directed towards the equilibrium position and is opposite in sign to x. A common slip: Assuming acceleration always opposes velocity. Instead, in SHM, acceleration points towards equilibrium, so it opposes velocity when moving away from equilibrium, but acts in the same direction when moving towards it. Exam tip: Sketch x–t, v–t and a–t on the same time axis so that maxima, minima and zero crossings line up; this makes the gradient relationships visible.
Key words
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Graphical interpretation of gradients
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Restoring force and acceleration towards equilibrium
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Phase relationships in simple harmonic motion
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See every definition
- Graphical interpretation of gradients
- The gradient of the displacement–time graph at a given instant gives the instantaneous velocity, and the gradient of the velocity–time graph at a given instant gives the instantaneous acceleration.
- Restoring force and acceleration towards equilibrium
- Acceleration is opposite in sign to displacement and always points towards the equilibrium position, opposing velocity only when moving away from equilibrium.
- Phase relationships in simple harmonic motion
- Where displacement is maximum the gradient is zero so velocity is zero, and where displacement crosses zero the gradient is steepest so velocity is maximum in magnitude.
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