Block on a Spring

Section 16.2 Block on a Spring

The angular frequency for a block of mass \(m\) attached to a massless spring with constant \(k\) is

\begin{equation*} \omega_s = \sqrt{\frac{k}{m}} \end{equation*}

The equation of motion for the block is given by

\begin{equation*} x(t) = x_{\text{max}} \cos(\omega_s t + \phi_i) \end{equation*}

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Derivation 16.2.3. Equation of Motion for a Block on a Spring.

Consider a spring with spring constant \(k\) that has one end held fixed in space and one end attached to a block of mass \(m\text{,}\) which rests on a frictionless surface. Assume the spring has negligible mass compared to the object’s mass \(m\) such that we can apply the massless spring approximation.

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The spring has equilibrium length \(x_0\) and is initially stretched and then released. You know from experience that the spring will undergo oscillatory motion—what is the equation of motion, the period, and the oscillation frequency for the spring? First draw a diagram of the situation and choose a coordinate system.

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If you place the origin of our coordinate system at equilibrium, the simplified spring force is a restoring force given by Hooke’s law:

\begin{equation*} \vec{F}(x) = -kx \hat{x} \end{equation*}

The negative sign tells you that the force always points in the opposite direction as the displacement. Applying the Law of Motion to the system and setting the force equal to the spring force

\begin{equation} ma_x = -kx \tag{16.2.1} \end{equation}

Using the relationship between position and acceleration

acceleration is the second derivative of position with respect to time

\begin{equation*} a_x(t) = \frac{d}{dt} v_x(t) = \frac{d^2}{dt^2}x(t) \end{equation*}

and equation (16.2.1) can be written in terms of position only

\begin{equation} \frac{d^2}{dt^2}x(t) = -\frac{k}{m} x(t)\tag{16.2.2} \end{equation}

Equation (16.2.2) is called the equation of motion for the mass on the spring. This is a second-order differential equation. Unlike algebraic functions whose solutions are numbers, solutions of differential equation are functions. This differential equation is one in which taking two derivatives of the function \(x(t)\) yields the original function multiplied by a negative constant.

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Often you can guess the solutions to differential equations. Since you already know the motion will be oscillatory from everyday experience, the natural guess for this physical situation is a position function that is oscillatory in time. Consider the position function

\begin{equation*} x(t) = x_{\text{max}} \cos(\omega t + \phi_i) \end{equation*}

Taking two derivatives in time

\begin{equation*} \frac{d^2}{dt^2}x(t) = - \omega^2 x_{\text{max}} \cos(\omega t + \phi_i) \end{equation*}

Plugging all this back into the equation of motion yields

\begin{equation*} - \omega^2 x_{\text{max}} \cos(\omega t + \phi_i) = -\frac{k}{m} x_{\text{max}} \cos(\omega t + \phi_i) \end{equation*}

From this you can determine the oscillation frequency of the spring system. Note that a factor of the position function can be canceled from each side of the equation. From this you can determine the angular frequency of the oscillation.

\begin{equation} \omega_s = \sqrt{\frac{k}{m}}\tag{16.2.3} \end{equation}

Since the oscillation frequency depends solely on the physical characteristics of the system (the stiffness of the spring and the mass of the object attached to the spring), the more specific label \(\omega_s\) is appropriate.

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Subsubsection Activities

Activity 16.2.1. Sense-making: Co-variational reasoning.

If you double the mass of the spring system, how does this affect the frequency of oscillation?

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If you double the amplitude of the oscillation, how does this affect the frequency of oscillation?

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If you double the stiffness of the spring, how does this affect the frequency of oscillation?

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If we took the spring system to another planet where the acceleration due to gravity at the surface of the planet was twice as large as on Earth, how would this affect the frequency of oscillation?

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Answer.

If \(m \rightarrow 2m \text{,}\) then \(\omega \rightarrow \frac{1}{\sqrt{2}}\omega_0\)

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If \(x_{\text{max}} \rightarrow 2x_{\text{max}} \text{,}\) then \(\omega = \omega_0 \text{.}\) This is an important point to note: The amplitude does not affect the oscillation frequency in simple harmonic motion.

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If \(k \rightarrow 2k \text{,}\) then \(\omega \rightarrow \sqrt{2}\omega_0 \)

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If \(g \rightarrow 2g \text{,}\) then \(\omega = \omega_0\text{.}\) The oscillation frequency does not depend on \(g\text{.}\)

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