From Velocity to Acceleration
A velocity that varies sinusoidally does not have constant acceleration. Its acceleration changes over time, and its sign indicates whether the velocity is increasing or decreasing. Differentiating the velocity function gives this changing acceleration directly.
The calculation involves more than differentiating sine or cosine: the frequency inside the trigonometric function contributes a factor, and the derivative of cosine has a negative sign. These details determine both the size and the timing of the acceleration.
The General Sinusoidal Form
A useful model for oscillating velocity is a sinusoid with a possible constant offset:
Here \(C\) is the velocity about which the motion oscillates, \(A\) is the velocity amplitude, \(\omega\) is the angular frequency, and \(\phi\) is the phase constant. The angle \(\omega t+\phi\) is the changing phase. A cosine model, \(v(t)=C+A\cos(\omega t+\phi)\), describes the same kind of oscillation with a different phase choice.
The period \(T\) is the time for the phase to increase by \(2\pi\). Since the phase increases at rate \(\omega\), the period is \(T=2\pi/\omega\). If the model is written using ordinary frequency \(f\), measured in cycles per unit time, then \(\omega=2\pi f\). Thus, differentiating a sinusoid written with \(f\) introduces the factor \(2\pi f\), not just \(f\).
Deriving the Acceleration Formula
The derivative formulas for sine and cosine, together with the chain rule for an affine input, give the acceleration. In the sine model, the inner function is \(\omega t+\phi\), whose derivative is \(\omega\).
To see why the first formula has these features, differentiate each part. The constant offset contributes zero. The outside sine becomes cosine, and the chain rule supplies the factor \(\omega\):
This also proves that the acceleration amplitude is \(A\omega\). A faster oscillation, with larger \(\omega\), produces a larger acceleration amplitude when the velocity amplitude \(A\) stays fixed. The offset \(C\) does not affect acceleration, because a constant velocity component has zero derivative.
Worked Examples
Worked Example: Differentiate a Sine Velocity
Suppose the velocity is \(v(t)=3+4\sin(2t-\pi/6)\). Find the acceleration and evaluate it at \(t=\pi/12\).
Differentiate the constant term and then apply the chain rule to the sine term. The derivative of the inner angle \(2t-\pi/6\) is \(2\):
At \(t=\pi/12\), the angle is \(2(\pi/12)-\pi/6=\pi/6-\pi/6=0\). Therefore, \(a(\pi/12)=8\cos(0)=8\). To check the substitution against the original velocity, \(v(\pi/12)=3+4\sin(0)=3\); the derivative formula at that same time gives \(8\), as calculated. The constant \(3\) contributes no acceleration.
Worked Example: Keep the Negative Sign for Cosine
Let \(v(t)=5\cos(3t+\pi/4)-2\). Find the acceleration and evaluate it at \(t=\pi/12\).
The derivative of cosine is negative sine. The inner angle \(3t+\pi/4\) has derivative \(3\), so both the negative sign and the factor \(3\) must appear:
At \(t=\pi/12\), the angle is \(3(\pi/12)+\pi/4=\pi/4+\pi/4=\pi/2\). Hence \(a(\pi/12)=-15\sin(\pi/2)=-15\). At this time the velocity is \(v(\pi/12)=5\cos(\pi/2)-2=-2\), which is its offset value. The acceleration formula gives the derivative at that time as \(-15\), confirming the evaluation.
Worked Example: Evaluate Acceleration at a General Time
Suppose \(v(t)=7+3\cos(\pi t/2-\pi/3)\). Find \(a(t)\), then find both \(v(1)\) and \(a(1)\).
The derivative of the inner angle \(\pi t/2-\pi/3\) is \(\pi/2\). Using the cosine derivative:
At \(t=1\), the angle is \(\pi/2-\pi/3=\pi/6\). Thus the velocity is
and the acceleration is
Substituting \(t=1\) into the derived acceleration formula gives exactly \(-3\pi/4\), so the evaluation is checked. The negative value means the velocity is decreasing at that instant.
Phase, Zeros, and Extreme Values
The phase relationship is visible by rewriting the acceleration in the same trigonometric form as the velocity. The identity \(\cos(\theta)=\sin(\theta+\pi/2)\) gives
when \(v(t)=C+A\sin(\omega t+\phi)\). The acceleration phase is ahead of the velocity phase by \(\pi/2\), which is one quarter of a full \(2\pi\) cycle. Since one full cycle takes time \(T\), a quarter-cycle corresponds to \(T/4\). In this sinusoidal model, acceleration reaches its maximum one quarter-period before velocity reaches its maximum.
There is also a direct way to interpret the relationship. Acceleration is the slope of the velocity graph:
- When acceleration is positive, velocity is increasing.
- When acceleration is negative, velocity is decreasing.
- When acceleration is zero, the velocity graph has a horizontal tangent. For a nonconstant sinusoid, this occurs at a maximum or minimum of velocity.
- The greatest positive or negative acceleration occurs as velocity crosses its midpoint \(C\), where the velocity is changing most rapidly.
For example, in the sine model, \(a(t)=0\) when \(\cos(\omega t+\phi)=0\). At those phases, the sine velocity is at one of its extreme values. When \(\cos(\omega t+\phi)=1\), acceleration is \(A\omega\), its maximum; when the cosine is \(-1\), acceleration is \(-A\omega\), its minimum. The same conclusions follow from the cosine model, with its corresponding phase shift.
A Common Sign and Frequency Pitfall
A frequent error is to differentiate cosine as positive sine, or to differentiate the outer trigonometric function but omit the derivative of its inner angle. For instance, the following calculation is wrong:
It has both the wrong sign and a missing factor of \(3\). The derivative of cosine is negative sine, and the chain rule multiplies by the derivative of \(3t+\pi/4\), which is \(3\). The corrected calculation is
A reliable check is to identify the derivative of the inner angle before simplifying. Also check the sign from the outer function: sine differentiates to cosine, while cosine differentiates to negative sine.
Why This Relationship Matters
Knowing the acceleration formula makes it possible to read how quickly an oscillating velocity changes and when that change reverses direction. The frequency factor matters: two velocity models with the same amplitude but different frequencies have different acceleration amplitudes. Their velocities may vary through the same range, but the faster oscillation changes velocity more sharply.
The phase relationship is equally useful. A velocity maximum is not an acceleration maximum; at a smooth maximum of velocity, acceleration is zero. The greatest positive acceleration occurs when velocity is passing upward through its midpoint, and the greatest negative acceleration occurs when velocity is passing downward through its midpoint. Keeping these relationships in mind provides a useful check on both a derivative formula and its interpretation.
Check Your Understanding
Work through these questions using the derivative formulas and phase relationships from this tutorial.
- For \(v(t)=2+6\sin(5t-\pi/3)\), find \(a(t)\) and its maximum possible value.
- For \(v(t)=4\cos(2t+\pi/6)\), identify the sign in the acceleration formula and state its amplitude.
- In a sine-form velocity model, what is the acceleration when velocity is at a maximum or minimum?
- How long is a quarter-cycle if the angular frequency is \(\omega=8\)?
- For \(v(t)=C+A\sin(\omega t+\phi)\), explain why changing \(C\) does not change acceleration.
