Topic 3 of 6
Connect velocity and acceleration
Velocity is the gradient of displacement against time, and acceleration is the gradient of velocity against time. Use those relationships to check both signs and the timing of maxima or zeros.
For ideal SHM with x = x0 sin(ωt), choose t = 0 at a rightward equilibrium crossing. Then:
a = -ω2x = -ω2x0 sin(ωt)
Here v0 is the velocity amplitude, or maximum speed. It is also the initial velocity only for this chosen phase. Displacement, velocity and acceleration are signed; their amplitudes are non-negative.
Read three aligned graphs
Use x0 = 0.0500 m, ω = 4.00 rad/s and T = 1.57080 s. The model is:
v = 0.200 cos(4.00t) m/s
a = -0.800 sin(4.00t) m/s2
Three quantities, one shared time scale
All three panels use exactly the same time axis, with T = π/2 s. The quarter-period labels are exact; the seconds beneath them are rounded. Each vertical axis has its own units and scale.
Displacement x
At t = 0 the cart crosses equilibrium to the right. Its next positive extreme is at T/4.
Velocity v
The velocity is the gradient of the displacement graph. Its sign tells the direction of motion.
Acceleration a
Acceleration is opposite in sign to displacement. It is zero at the centre and largest in magnitude at either extreme.
| Time | x / m | v / (m/s) | a / (m/s2) |
|---|---|---|---|
| 00 s | 0 | +0.200 | 0 |
| T/40.392699 s | +0.0500 | 0 | -0.800 |
| T/20.785398 s | 0 | -0.200 | 0 |
| 3T/41.17810 s | -0.0500 | 0 | +0.800 |
| T1.57080 s | 0 | +0.200 | 0 |
At t = 0 the cart crosses the centre to the right. At T/4 it turns at the positive extreme; at T/2 it crosses the centre to the left. At 3T/4 it turns at the negative extreme, and at T the original state repeats.
At the centre: speed is greatest and restoring acceleration is zero. At either extreme: velocity is momentarily zero, but acceleration has greatest magnitude towards equilibrium. Zero instantaneous velocity is therefore not enough to identify equilibrium.
Comparing corresponding stages of the sinusoidal traces, velocity leads displacement by π/2 rad. Acceleration and displacement are in antiphase, separated by π rad: when x is most positive, a is most negative. Their different units do not share one common vertical scale.
When x is positive, acceleration is negative on both passes. It slows a cart moving right, but speeds a cart returning left. Acceleration is set by displacement, not by a rule that it must always oppose velocity.
Find velocity from position and direction
Using sin2(ωt) + cos2(ωt) = 1 to eliminate time gives:
v = ±ω√(x02 - x2)
The square root gives speed magnitude. Select + for motion towards positive x and - for motion towards negative x. The model requires |x| ≤ x0; a position beyond the stated amplitude does not belong to this motion.
Worked signed velocity
The same displacement can occur on two passes
At x = +0.0300 m with x0 = 0.0500 m and ω = 4.00 rad/s:
= 0.160 m/s
a = -16.0(0.0300) = -0.480 m/s2
If returning towards equilibrium, v = -0.160 m/s. If moving outward to the right, v = +0.160 m/s. The acceleration is -0.480 m/s2 in both cases.
For the 0.400 kg cart, the resultant force is ma = -0.192 N in either state. Position alone fixes the restoring acceleration, but not which of the two velocity directions applies.
Optional check An SHM model has amplitude 0.0500 m and angular frequency 4.00 rad/s. At x = +0.0300 m, the body is returning towards equilibrium. What are its velocity and acceleration?
Estimate rates from a position record
The following five positions are generated from the same sine model and rounded to 0.00001 m. They illustrate the calculation method; they are not observations that independently test the model.
| Time t / s | Displacement x / m |
|---|---|
| 0.00 | 0.00000 |
| 0.10 | 0.01947 |
| 0.20 | 0.03587 |
| 0.30 | 0.04660 |
| 0.40 | 0.04998 |
For each adjacent interval, calculate Δx/Δt. The four average velocities are 0.1947, 0.1640, 0.1073 and 0.0338 m/s, assigned approximately to interval midpoints 0.05, 0.15, 0.25 and 0.35 s. Keep these derived values separate from the original positions.
For example, the velocity over 0.10 to 0.20 s is (0.03587 - 0.01947)/0.10 = 0.1640 m/s. The next interval gives (0.04660 - 0.03587)/0.10 = 0.1073 m/s.
= (0.1073 - 0.1640)/(0.25 - 0.15)
= -0.567 m/s2
Pair this estimate with x = 0.03587 m at 0.20 s. The exact sine model gives about -0.573885 m/s2 there. Finite intervals and position rounding explain why the estimate differs; it is not the exact instantaneous acceleration.
To investigate real motion, apply the method to an actual calibrated x-t record spanning both sides of equilibrium. Compare the observed displacement pattern with a sinusoid, then plot acceleration estimates against displacements at corresponding times. A roughly straight a-x relationship through the origin with negative gradient supports the SHM model over the tested range. Its gradient estimates -ω2, which can be compared with the independently timed period.
The few positive-side estimates in this generated illustration do not establish that relationship experimentally and do not need a fitted line. For real records, check calibration, equilibrium zero, frame timing, changing amplitude and possible nonlinearity.
Shorter intervals can improve a finite-gradient approximation, but differences then become smaller relative to measurement resolution. Estimating rates amplifies noise, especially when velocity estimates are differenced again. Choose intervals that resolve the motion while still giving reliable position and timing differences; smaller is not automatically better.