From Position to Velocity
A sinusoidal position model describes motion that repeats in a regular cycle. Its graph oscillates around an equilibrium position: the displacement is largest at the ends of the motion and passes through equilibrium between them. Differentiating position gives velocity, which describes both how quickly the position changes and in which direction.
The derivative changes the form of the sinusoid in a systematic way. Its amplitude depends on both the displacement amplitude and the angular frequency, while its phase shifts by a quarter-cycle. Understanding these features lets us interpret the velocity model, not just calculate it.
The angle \(\omega t+\phi\) is the phase at time \(t\). Angular frequency measures how quickly that phase changes. If a motion completes \(f\) cycles per unit time, then \(\omega=2\pi f\). The period, or time for one full cycle, is \(T=2\pi/\omega\).
The Velocity Formula
Differentiate the position model using the chain rule. The derivative of the constant equilibrium position is zero, and differentiating the sine requires multiplying by the derivative of its input, \(\omega t+\phi\).
This formula gives the velocity at every time. Since cosine ranges from \(-1\) to \(1\), velocity ranges from \(-A\omega\) to \(A\omega\). Thus, the velocity amplitude—the maximum magnitude of velocity—is \(A\omega\). The velocity has the same period \(T\) as the position because its phase still has angular frequency \(\omega\).
The velocity can also be written in sine form. Since \(\cos\theta=\sin(\theta+\pi/2)\), we have
So in this sine representation, velocity is shifted forward in phase by \(\pi/2\), which is one quarter of a cycle. This means velocity reaches its positive maximum when the position passes through equilibrium in the increasing direction. It reaches its negative maximum when position passes through equilibrium in the decreasing direction. At a position extreme, the velocity is zero.
Why the Formula Has These Features
The derivative formula follows directly from the chain rule and the derivative of sine. The interpretations of amplitude and phase follow from basic properties of cosine. Here is the argument in full.
For \(x(t)=x_0+A\sin(\omega t+\phi)\), the constant \(x_0\) differentiates to zero. The chain rule gives \(x'(t)=A\omega\cos(\omega t+\phi)\).
Because \(-1\leq\cos(\omega t+\phi)\leq1\), multiplication by the positive constant \(A\omega\) gives \(-A\omega\leq v(t)\leq A\omega\). Therefore the maximum speed is \(A\omega\).
The identity \(\cos\theta=\sin(\theta+\pi/2)\) rewrites the velocity with phase \(\phi+\pi/2\). The phase still increases by \(2\pi\) when time increases by \(2\pi/\omega\), so the velocity period is \(2\pi/\omega\), the same as the position period.
This also explains the effect of changing the model parameters. Increasing \(A\) scales the maximum displacement and the maximum velocity by the same factor. Increasing \(\omega\) makes the cycle faster and also multiplies the velocity amplitude by that factor. Changing \(\phi\) shifts the timing of the cycle but does not change either amplitude or period. Changing \(x_0\) moves the position up or down without changing its rate of change.
Worked Examples
Worked Example: A Sine-Model Position
A particle’s position in metres is \(x(t)=2+0.4\sin(5t-\pi/6)\), where \(t\) is measured in seconds. Find its velocity function, velocity amplitude, and period.
Differentiate term by term. The derivative of \(2\) is zero. The inner angle \(5t-\pi/6\) has derivative \(5\), so the chain rule gives
The velocity amplitude is \(A\omega=0.4(5)=2\ \text{m/s}\). The period is \(T=2\pi/5\ \text{s}\). The equilibrium position \(2\ \text{m}\) contributes no velocity.
Check: Differentiating the original position gives \(x'(t)=0+0.4(5)\cos(5t-\pi/6)=2\cos(5t-\pi/6)\ \text{m/s}\), exactly the stated velocity. At \(t=\pi/30\ \text{s}\), the angle is \(5(\pi/30)-\pi/6=0\), so \(v(\pi/30)=2\cos(0)=2\ \text{m/s}\), the positive maximum.
Worked Example: A Cosine-Model Position
Suppose \(x(t)=1.5-0.8\cos(4t+\pi/3)\) metres. Find velocity and express it both as a sine and a cosine.
Let \(\theta=4t+\pi/3\), so \(d\theta/dt=4\). Differentiating gives
Since \(\sin\theta=\cos(\theta-\pi/2)\), an equivalent form is \(v(t)=3.2\cos(4t+\pi/3-\pi/2)=3.2\cos(4t-\pi/6)\ \text{m/s}\). The velocity amplitude is \(0.8(4)=3.2\ \text{m/s}\), and the period is \(2\pi/4=\pi/2\ \text{s}\).
Check: Differentiating the original expression, the derivative of \(-0.8\cos\theta\) is \(-0.8(-\sin\theta)(4)=3.2\sin\theta\), as stated. The two velocity forms agree because \(\cos(\theta-\pi/2)=\sin\theta\); no extra sign or frequency factor is introduced by this identity.
Worked Example: Using Frequency in Cycles per Second
A position model is \(x(t)=10+3\sin\bigl((\pi/2)t-\pi/2\bigr)\) centimetres. Find the velocity and interpret it at \(t=0\) and \(t=1\) second. The angular frequency is \(\omega=\pi/2\) radians per second, corresponding to \(f=\omega/(2\pi)=1/4\) cycle per second.
Differentiate the model, including the derivative \(\pi/2\) of the inner angle:
At \(t=0\), the phase is \(-\pi/2\). Thus \(x(0)=10+3\sin(-\pi/2)=10-3=7\ \text{cm}\), and \(v(0)=(3\pi/2)\cos(-\pi/2)=0\ \text{cm/s}\). At \(t=1\), the phase is \(0\), so \(x(1)=10+3\sin(0)=10\ \text{cm}\) and \(v(1)=(3\pi/2)\cos(0)=3\pi/2\ \text{cm/s}\), approximately \(4.71\ \text{cm/s}\) to two decimal places. The particle is at equilibrium at \(t=1\) and moving in the positive direction at its maximum velocity.
Check: Differentiating the original model yields \(3(\pi/2)\cos((\pi/2)t-\pi/2)\), which is the velocity formula above. The period is \(1/f=4\ \text{s}\), also \(2\pi/(\pi/2)=4\ \text{s}\).
Common Pitfall: Omitting the Frequency Factor
A frequent error is differentiating the outer sine but forgetting to differentiate its input. For example, a mistaken calculation for \(x(t)=3\sin(4t)\) might be
This omits the factor \(4\), which comes from the chain rule. The correct calculation is
As a check at \(t=0\), the correct derivative is \(12\cos(0)=12\), whereas the incorrect expression gives \(3\). A quick test is to identify the derivative of the inner angle before finishing: the velocity amplitude must be displacement amplitude times angular frequency, not just displacement amplitude.
Interpreting the Resulting Motion
Velocity tells more than the rate at which position changes. Its sign identifies direction: positive velocity means position is increasing, and negative velocity means position is decreasing. Its magnitude indicates how rapidly the position changes. In sinusoidal motion, these features repeat with the same period as position, but their timing differs by a quarter-cycle.
At a displacement extreme, the position momentarily stops changing direction, so velocity is zero. As the particle passes through equilibrium, the magnitude of velocity is largest. The phase determines when these events occur. Therefore, a velocity formula should be read through its amplitude, period, phase, and sign—not as an isolated derivative expression.
When checking a result, differentiate the original position model and compare the full expression, including its sign and chain-rule factor. If a velocity has been rewritten using a trigonometric identity, verify that identity directly. For instance, \(A\omega\cos\theta=A\omega\sin(\theta+\pi/2)\). This confirms the equivalent phase form without confusing a velocity expression with a derivative of velocity.
Check Your Understanding
Use the position-to-velocity relationship to answer each question.
- If \(x(t)=5+2\sin(3t+\pi/4)\), what is \(v(t)\), and what is its velocity amplitude?
- For \(x(t)=4\cos(2t)\), find velocity in both sine and cosine form.
- For a sine-model position, at what phase is velocity zero? At what phase is it at its positive maximum?
- How does doubling angular frequency affect velocity amplitude when displacement amplitude is unchanged?
- Why does adding a constant to a position model not change its velocity?
