Velocity and acceleration in SHM
Velocity in SHM
Back to the reference circle. P moves round a circle of radius A with constant angular speed ω, so its speed is v equals ωA, along the tangent.
P′, the foot of P on the x-axis, moves in SHM. Its velocity is just the x-component of P’s velocity. Watch the green velocity arrow on P cast its shadow on the axis.
When the phase is ωt plus φ, the tangent velocity of P has an x-component of minus ωA sin of ωt plus φ. So the velocity in SHM is v of t equals minus ωA sin of ωt plus φ.
The minus sign makes sense. When P is in the upper half of the circle, it is moving to the left, so P′ moves towards minus x.
You get the same result by differentiating x equals A cos of ωt plus φ with respect to time. The reference circle just lets you see it.
At the mean position, x is zero and the sine is plus or minus one, so the speed is greatest: ωA. At the extreme positions, x is plus or minus A, the sine is zero, and the particle is momentarily at rest.
Using sin squared plus cos squared equals one, we can write the speed at any displacement: v equals ω times the square root of A squared minus x squared.
Worked example: amplitude 10 cm and ω equals 10 radians per second. The maximum speed is ωA, 10 times 0.1, which is 1 metre per second. At x equals 6 cm, v is 10 times the square root of 0.01 minus 0.0036, which is 0.8 metres per second.
Acceleration and phase
P moves in a circle at constant speed, so it has a centripetal acceleration ω squared A, always pointing towards the centre. Watch the amber arrow.
The acceleration of P′ is the x-component of this arrow: minus ω squared A cos of ωt plus φ. But A cos of ωt plus φ is just x. So a equals minus ω squared x.
This is the heart of SHM. The acceleration is proportional to the displacement and always points towards the mean position. Pull the particle right, and it accelerates left.
At the mean position, the acceleration is zero. At the extremes, it is largest, ω squared A, just where the velocity is zero. Being at rest does not mean having no acceleration.
Now stack the three graphs: x, v and a against time. The velocity curve leads the displacement by a quarter of a period, a phase of π by 2.
The acceleration leads the velocity by another π by 2. So acceleration and displacement are always opposite: a phase difference of π.
Worked example: x equals 5 cos of 2πt plus π by 4, in metres. At t equals 1.5 seconds, the phase is 3π plus π by 4. So x equals 5 times minus 0.707, which is minus 3.54 metres.
The velocity is minus 10π times sin of 3π plus π by 4, which comes to plus 22 metres per second. The acceleration is minus 4π squared times x, about plus 140 metres per second squared.
Notice the signs: the particle is on the negative side, but its acceleration is positive, pointing back towards the centre, just as a equals minus ω squared x demands.
Try it: x, v and a together
- Press Play. The green and amber arrows on P′ are the shadows of P’s velocity and acceleration.
- Pause when x is at +A. What are v and a at that moment?
- Pause when P′ crosses the centre. Which arrow has vanished?
- Compare the peaks of the three graphs: each is a quarter period apart.
Through the centre
At the instant a particle in SHM passes through its mean position:
- its velocity is zero and its acceleration is maximum
- its speed is maximum and its acceleration is zero
- both its speed and its acceleration are maximum
- both its velocity and its acceleration are zero
Answer
B. At the mean position x = 0, so a = −ω²x = 0. The speed v = ω√(A² − x²) is largest there: ωA. At the extremes it is the other way round: v = 0 and |a| = ω²A.
Maximum speed and acceleration
A particle performs SHM with amplitude 4 cm and period 0.5 s. What are its maximum speed and its maximum acceleration?
- 50 cm/s and 6.3 m/s²
- 25 cm/s and 1.6 m/s²
- 8 cm/s and 0.32 m/s²
- 50 cm/s and 0.5 m/s²
Answer
A. ω = 2π/T = 2π/0.5 = 4π ≈ 12.6 rad/s. Maximum speed = ωA = 12.6 × 0.04 ≈ 0.50 m/s. Maximum acceleration = ω²A = 158 × 0.04 ≈ 6.3 m/s². Using ω = 2π (forgetting to divide by T) gives the second option.
Summary
- Displacement:
x = A cos(ωt + φ) - Velocity:
v = −ωA sin(ωt + φ) - Speed at displacement x:
v = ω√(A² − x²) - Acceleration:
a = −ω²A cos(ωt + φ) = −ω²x - Maximum values:
vₘ = ωA, aₘ = ω²A
- The velocity of the SHM particle is the projection of the velocity of the reference point P on the circle.
- Speed is greatest at the mean position and zero at the extremes. Acceleration is zero at the mean position and greatest at the extremes.
- a = −ω²x: the acceleration is proportional to the displacement and always directed towards the mean position.
- v leads x by π/2, and a leads v by π/2, so a and x are opposite in phase.