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Learning Introductory Physics with Activities

Paul J. Emigh, Rebecka Tumblin, Kathryn Hadley, Danielle Skinner

Section 6.3 Kinematics Relations

You can use differential calculus to find kinematic relationships for an object undergoing oscillatory motion. The time derivative of position yields the velocity of the system.
\begin{equation*} \frac{d}{dt} \vec{x}(t) = \vec{v}(t) \end{equation*}
Taking the time derivative of the position function \(\vec{x}(t) = \vec{x}_o + A \cos{(\omega t + \phi_o)}\) yields the velocity function \(\vec{v}(t)\)
\begin{equation} \vec{v}(t) = - \omega A \sin(\omega t + \phi_o)\hat{x} \tag{6.3.1} \end{equation}
As the object passes through the equilibrium position located at \(\vec{x} = 0 \text{,}\) the speed (magnitude of the velocity) is maximum. When the object is maximally displaced from equilibrium, it temporarily comes to rest before changing its direction. Recall that a maxima or minima in the position as a function of time graph corresponds to a turning point in the motion. Therefore, when the object is located at \(x= \pm A\) the instantaneous velocity of the object is zero.
Figure 6.3.1. The position and velocity as a function of time for an object in Simple Harmonic Motion. The graphs are aligned such that the time axes in both graphs correspond to the same point in time.

Exercises Activities

1. Physical Units.

Looking at equation (6.3.1) what are the SI units of the quantity \(\omega A\text{?}\) What is the physical significance of this quantity?
Answer.
The equation as a whole is a velocity and has dimension of length/time with SI units of meters/second. Trigonometric functions are dimensionless. Therefore, the quantity \(\omega A\) must have dimension of length/time for the equation to be true. Note that the SI units of \(\omega\) are \(rad/s\text{.}\) While radians are an SI unit, they are dimensionless. The physical significance of this quantity is that it corresponds to the maximum speed \(v_{max} = \omega A \) of the object. This allows us to re-write equation Figure 6.3.1 as follows
\begin{equation} \vec{v}(t) = - v_{max} \sin(\omega t + \phi_0) \hat{x} \tag{6.3.2} \end{equation}

2. Exploring Acceleration.

Derive the acceleration of an object in simple harmonic motion using the methods discussed in this section.
Answer.
Looking at equation Figure 6.3.1 we take the time derivative of velocity determine the acceleration of the system.
\begin{equation} \vec{a}(t) = - \omega^2 A \cos(\omega t + \phi_0) \hat{x} \tag{6.3.3} \end{equation}
Note that the quantity \(\omega^2 A = a_{max}\) represents the magnitude of the acceleration of the object when \(\vec{x}= \pm A \hat{x}\text{.}\)

3. Exploring Acceleration Graphically.

Referencing figure Figure 6.3.1, sketch the accompanying \(a_x(t)\) function.

References References

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