3.4: Analyzing differential equations- The spring-mass system

Estimating derivatives reduces differentiation to division (Section 3.3); it thereby reduces differential equations to algebraic equations.

To produce an example equation to analyze, connect a block of mass m to an ideal spring with spring constant (stiffness) \(k\), pull the block a distance \(x_{0}\) to the right relative to the equilibrium position \(x = 0\), and release it at time \(t = 0\). The block oscillates back and forth, its position \(x\) described by the ideal-spring differential equation

\[m\frac{d^{2}x}{dt^{2}}+kx = 0. \label{3.21} \]

Let’s approximate the equation and thereby estimate the oscillation frequency.

Checking dimensions

Upon seeing any equation, first check its dimensions (Chapter 1). If all terms do not have identical dimensions, the equation is not worth solving a great savings of effort. If the dimensions match, the check has prompted reflection on the meaning of the terms; this reflection helps prepare for solving the equation and for understanding any solution.

Question

What are the dimensions of the two terms in the spring equation?

Look first at the simple second term \(kx\). It arises from Hooke’s law, which says that an ideal spring exerts a force \(kx\) where \(x\) is the extension of the spring relative to its equilibrium length. Thus the second term \(kx\) is a force. Is the first term also a force?

The first term m(d\(^{2}\)x/dt\(^{2}\)) contains the second derivative \(d^{2}x/dt^{2}\), which is familiar as an acceleration. Many differential equations, however, contain unfamiliar derivatives. The Navier–Stokes equations of fluid mechanics (Section 2.4),

\[\frac{\partial \mathbf{v}}{\partial t}+(\mathbf{v} \cdot \boldsymbol{\nabla}) \mathbf{v}=-\frac{1}{\rho} \boldsymbol{\nabla} p+v \boldsymbol{\nabla}^{2} \mathbf{v}\label{3.22} \]

contain two strange derivatives: \((v·∇)v\) and \(∇^{2}v\). What are the dimensions of those terms?

To practice for later handling such complicated terms, let’s now find the dimensions of \(d^{2}x/dt^{2}\) by hand. Because \(d^{2}x/dt^{2}\) contains two exponents of 2, and \(x\) is length and t is time, \(d^{2}x/dt^{2}\) might plausibly have dimensions of \(L^{2}T^{2}\).

Question

Are \(L^{2}T^{-2}\) the correct dimensions?

To decide, use the idea from Section 1.3.2 that the differential symbol d means “a little bit of.” The numerator \(d^{2}x\), meaning \(d\) of dx, is “a little bit of a little bit of x.” Thus, it is a length. The denominator \(dt^{2}\) could plausibly mean \((dt)^{2}\) or \(d(t^{2}\)). [It turns out to mean \((dt)^{2})\). In either case, its dimensions are \(T^{2}\). Therefore, the dimensions of the second derivative are \(LT^{-2}\):

\[[\frac{d^{2}}{dt^{2}}] = LT^{-2}\label{3.23} \]

This combination is an acceleration, so the spring equation’s first term \(m(d^{2}x/dt^{2})\) is mass times acceleration giving it the same dimensions as the \(kx\) term.

Problem 3.18 Dimensions of spring constant

What are the dimensions of the spring constant \(k\)?

Estimating the magnitudes of the terms

The spring equation passes the dimensions test, so it is worth analyzing to find the oscillation frequency. The method is to replace each term with its approximate magnitude. These replacements will turn a complicated differential equation into a simple algebraic equation for the frequency.

To approximate the first term \(m(d^{2}x/dt^{2})\), use the significant-change approximation (Section 3.3.3) to estimate the magnitude of the acceleration \(d^{2}x/dt^{2}\).

\[\frac{d^{2}x}{dt^{2}} ∼ \frac{\text{ significant Δx }}{\text{ (Δt that produces a significant Δx m)}^{2}}. \label{3.24} \]

Problem 3.19 Explaining the exponents

The numerator contains only the first power of \(Δx\), whereas the denominator contains the second power of \(Δt\). How can that discrepancy be correct?

To evaluate this approximate acceleration, first decide on a significant \(Δx\) on what constitutes a significant change in the mass’s position.

The mass moves between the points \(x = −x_{0}\) and \(x = +x_{0}\), so a significant change in position should be a significant fraction of the peak-to-peak amplitude \(2x_{0}\). The simplest choice is \(Δx = x_{0}\). Now estimate \(Δt\): the time for the block to move a distance comparable to \(Δx\). This time called the characteristic time of the system is related to the oscillation period T. During one period, the mass moves back and forth and travels a distance \(4x_{0}\) much farther than \(x_{0}\). If \(Δt\) were, say, \(T/4\) or \(T/2\pi\), then in the time \(Δt\) the mass would travel a distance comparable to \(x_{0}\). Those choices for \(Δt\) have a natural interpretation as being approximately \(1/ω\), where the angular frequency \(ω\) is connected to the period by the definition \(ω ≡ 2\pi/T\). With the preceding choices for \(Δx\) and \(Δt\), the \(m(d^{2}x/dt^{2})\) term is roughly \(mx_{0}ω^{2}\).

Question

What does “is roughly” mean?

The phrase cannot mean that \(mx_{0}ω^{2}\) and \(m(d^{2}x/dt^{2})\) are within, say, a factor of 2, because \(m(d^{2}x/dt^{2})\) varies and \(mx_{0}/τ^{2}\) is constant. Rather, “is roughly” means that a typical or characteristic magnitude of \(m(d^{2}x/dt^{2})\) for example, its root-mean-square value is comparable to \(mx_{0}ω^{2}\). Let’s include this meaning within the twiddle notation ∼. Then the typical-magnitude estimate can be written

With the same meaning of “is roughly”, namely that the typical magnitudes are comparable, the spring equation’s second term kx is roughly \(kx_{0}\). The two terms must add to zero a consequence of the spring equation

\[m\frac{d^{2}x}{dt^{2}} + kx = 0. \label{3.26} \]

Therefore, the magnitudes of the two terms are comparable:

\[mx_{0}w^{2} ∼ kx_{0}. \label{3.27} \]

The amplitude \(x_{0}\) divides out! With \(x_{0}\) gone, the frequency ω and oscillation period \(T = 2\pi/ω\) are independent of amplitude. [This reasoning uses several approximations, but this conclusion is exact (Problem 3.20).] The approximated angular frequency \(ω\) is then \(\sqrt{k/m}\).

For comparison, the exact solution of the spring differential equation is, from Problem 3.22,

\[x = x_{0} cos ωt, \label{3.28} \]

where \(ω\) is \(\sqrt{k/m}\). The approximated angular frequency is also exact!

Multiple problems

Problem 3.20 Amplitude independence

Use dimensional analysis to show that the angular frequency \(ω\) cannot depend on the amplitude \(x_{0}\).

Problem 3.21 Checking dimensions in the alleged solution

What are the dimensions of \(ωt\)? What are the dimensions of \(\cos ωt\)? Check the dimensions of the proposed solution \(x = x \cos ωt\), and the dimensions of the proposed period \(2\pi \sqrt{m/k}\).

Problem 3.22 Verification

Show that \(x = x_{0} \cos ωt\) with \(ω = \sqrt{k/m}\) solves the spring differential equation

\[m\frac{d^{2}x}{dt^{2}}+kx = 0. \label{3.29} \]

Meaning of the Reynolds number

As a further example of lumping in particular, of the significant-change approximation let’s analyze the Navier–Stokes equations introduced in Section 2.4,

\[\frac{\partial \mathbf{v}}{\partial t}+(\mathbf{v} \cdot \boldsymbol{\nabla}) \mathbf{v}=-\frac{1}{\rho} \boldsymbol{\nabla} p+v \boldsymbol{\nabla}^{2} \mathbf{v} \label{3.30} \]

and extract from them a physical meaning for the Reynolds number \(rv/ν\).

To do so, we estimate the typical magnitude of the inertial term \((v·∇)v\) and of the viscous term \(ν∇^{2}v\).

Question

What is the typical magnitude of the inertial term?

The inertial term \((v·∇)v\) contains the spatial derivative \(∇v\). According to the significant-change approximation (Section 3.3.3), the derivative \(∇v\) is roughly the ratio

\[\frac{\text{ significant change in flow velocity }}{\text{ distance over which flow velocity changes significantly }}. \label{3.31} \]

The flow velocity (the velocity of the air) is nearly zero far from the cone and is comparable to v near the cone (which is moving at speed v). Therefore, v, or a reasonable fraction of \(v\), constitutes a significant change in flow velocity. This speed change happens over a distance comparable to the size of the cone: Several cone lengths away, the air hardly knows about the falling cone. Thus \(∇v ∼ v/r\). The inertial term \((v·∇)v\) contains a second factor of \(v\), so \((v·∇)v\) is roughly \(v^{2}/r\).

Question

What is the typical magnitude of the viscous term?

The viscous term \(ν∇^{2}v\) contains two spatial derivatives of \(v\).

Because each spatial derivative contributes a factor of \(1/r\) to the typical magnitude, \(ν∇^{2}v\) is roughly \(νv/r^{2}\). The ratio of the inertial term to the viscous term is then roughly \((v^{2}/r)/(νv/r^{2})\). This ratio simplifies to rv/ν the familiar, dimensionless, Reynolds number.

Thus, the Reynolds number measures the importance of viscosity. When \(Re\) ≫ 1, the viscous term is small, and viscosity has a negligible effect. It cannot prevent nearby pieces of fluid from acquiring significantly different velocities, and the flow becomes turbulent. When \(Re\) ≪ 1, the viscous term is large, and viscosity is the dominant physical effect. The flow oozes, as when pouring cold honey.