How do you find velocity from a displacement–time graph?

It is the gradient (of the tangent) of the displacement–time graph.

How do you find acceleration from a velocity–time graph?

It is the gradient (of the tangent) of the velocity–time graph.

What is the velocity when the displacement is at its maximum?

Zero, because the displacement graph is flat at its peak.

Where is the speed greatest on a displacement–time graph?

Where the graph crosses zero (the equilibrium position), because the gradient is steepest.

What shape is the velocity–time graph for SHM starting from x = +A?

A negative sine curve, v = −Aω sin ωt.

What shape is the acceleration–time graph for SHM starting from x = +A?

A negative cosine curve, a = −Aω² cos ωt.

What is the maximum acceleration in SHM?

Aω², at the maximum displacement.

What is the phase relationship between acceleration and displacement graphs?

They are in antiphase (half a cycle apart), because a = −ω²x.

How can the period be found from a displacement–time graph?

From the time between two successive peaks (or troughs); then ω = 2π/T.

Displacement, velocity and acceleration graphs for SHMEdexcel A-Level Physics: Flashcards

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What shape is the displacement–time graph for SHM starting from maximum displacement?

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What shape is the displacement–time graph for SHM starting from maximum displacement?: A cosine curve, x = A cos ωt.
How do you find velocity from a displacement–time graph?: It is the gradient (of the tangent) of the displacement–time graph.
How do you find acceleration from a velocity–time graph?: It is the gradient (of the tangent) of the velocity–time graph.
What is the velocity when the displacement is at its maximum?: Zero, because the displacement graph is flat at its peak.
Where is the speed greatest on a displacement–time graph?: Where the graph crosses zero (the equilibrium position), because the gradient is steepest.
What shape is the velocity–time graph for SHM starting from x = +A?: A negative sine curve, v = −Aω sin ωt.
What is the maximum value of the velocity in SHM?: Aω.
What shape is the acceleration–time graph for SHM starting from x = +A?: A negative cosine curve, a = −Aω² cos ωt.
What is the maximum acceleration in SHM?: Aω², at the maximum displacement.
What is the phase relationship between acceleration and displacement graphs?: They are in antiphase (half a cycle apart), because a = −ω²x.
How far is the velocity graph out of phase with the displacement graph?: A quarter of a cycle (T/4).
How can the period be found from a displacement–time graph?: From the time between two successive peaks (or troughs); then ω = 2π/T.

Exam questions on Displacement, velocity and acceleration graphs for SHM

A trolley on a horizontal spring oscillates with simple harmonic motion. Its amplitude is 0.080 m and its period is 0.50 s. At t = 0 it is at its maximum positive displacement, so its displacement is x = 0.080 cos ωt. A student plots graphs of its displacement, velocity and acceleration against time.
Calculate the maximum acceleration of the trolley.2 marks
A mass on a vertical spring is released from rest at its maximum upward displacement of 0.050 m at t = 0. Upward is taken as positive, and the period of the oscillation is 0.80 s. A graph of its velocity against time is plotted.
Calculate the maximum speed of the mass and state the first time at which it occurs.2 marks
The displacement x of a pendulum bob from its equilibrium position is given by x = 0.060 cos(πt), where x is in metres and t is in seconds, so the period is 2.0 s. A graph of displacement against time is plotted for one complete cycle.
Calculate the velocity of the bob at t = 0.50 s and explain how this value could be found from the displacement–time graph.3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).