You step on your bathroom scales to find your body mass. The idea is that gravity acts on your mass to exert a force that depresses a platform. The depression of the platform depends on the force and, hence, on your body mass. So, to find your body mass, you need to measure the depression of the platform.

Previously the platform would have been mounted on springs but now it is more likely to be a sensor (post 18.9), in which an upper surface deforms (perhaps not enough for you to see), but the principle remains the same.

So the initial height of the platform, h, decreases to a new value. Ideally you might expect this to happen instantaneously, so that your scales would give you an immediate value for your body mass. But this doesn’t happen. The purpose of this post is to answer the question – why not?

Application of the force leads to acceleration of the platform with you standing on it. According to Newton’s first law of motion, we now have an accelerating mass that would continue moving at a constant speed for ever. But this doesn’t happen because components in the system have stiffness (like a spring, storing some of the energy of motion) and exert a drag force (like a dashpot, dissipating energy). We now have a second-order system that we will assume is described by a linear second-order differential equation; this is a reasonable assumption because the components will have been selected so that their properties don’t change during normal function. The motion of this system is described by

where x is the displacement of the platform, from its final position, at time t, k is the stiffness of the system, M(dx/dt) is the drag force exerted by the system and m is the mass of the moving parts. All this is explained in post 24.5.

If you compare this equation of motion with equation 1a of post 24.5, you will see that the force, whose modulus is F, exerted by gravity on your body mass, does not appear. This is because it has already initiated the motion creating a force m(d²x/dt²) within the total system. We now have a system that is moving, independent of its previous history, and is being brought to rest by stiffness and drag. This system is analogous to a pendulum that is displaced from its vertical rest position by a force. The initial motion of the pendulum is towards its rest position. But, once it has been displaced, its motion is independent of the force that initially displaced it.

We need to find an expression for x; notice that it must be similar to its first and second derivative if equation 1 equals zero. So we’ll try an exponential function of the form

x = Ae^at                     (2)

where A and a are unknown constants, as a trial solution so that

dx/dt = Aae^at    and   d²x/dt² = Aa²e^at.

Then equation 1 becomes

A(k + Ma + ma²)e^at = 0.

Since e^at cannot equal zero, the possible solutions to this equation are

A = 0   and   (k + Ma + ma²) = 0.

The first solution is of no interest to us because when A = 0, x = 0 and nothing moves. The second solution (in brackets) is a quadratic equation with two solutions

as described in post 18.5.

We also need to find a value for A. Notice that when t = 0, x = h. Substituting this boundary condition into equation 2 gives

x₀ = h.                    (4)

We are now going to consider three possible cases for the behaviour of our scales depending on the relative values of M² and 4km.

Case 1   M²/4km = 1

In this case, the expression under the square root sign of equation 3 is zero. Then, from equations 2, 3 and 4,

Here exp(y) is another way of writing e^y.

This is called critical damping.

Case 2   M²/4km > 1

Now the expression under the square root sign of equation 3 is a real number. In appendix 1, I show that now

Here cosh(y) is the hyperbolic cosine of y, defined in post 22.8.

This is called overdamping.

Case 3   M²/4km < 1

Now the expression under the square root sign of equation 3 is an imaginary number. In appendix 2, I show that now

This is called underdamping.

Comparison of the three cases

To compare the three cases, I define

The reason is to reduce the number of symbols in our equations. The table below then summarises all the results so far.

The picture above shows how x varies for each of the three cases. I have assigned h = 1 and p = 1 for all cases. But for case 2 (overdamping) I have assigned q = 0.8 and for case 3 (underdamping) q = 3; this is to obtain numbers that can be plotted on the same graph.

For case 1 (critical damping), the graph is a smooth curve showing x moving to its final value. In case 2, x approaches its final value of zero more slowly. And in case 3, x oscillates as it approaches zero; post 24.6 shows this is the behaviour of a damped simple harmonic oscillator. And, the greater the value of p, the greater the amplitude of the oscillations; as p approaches 1, the behaviour more closely resembles that of a critically damped system.

Our bathroom scales aren’t critically damped because to ensure that p² = 1 (that is M²/4km = 1), we need to know the value of M. But M is unknown – the purpose of the scales is to measure it! If they were overdamped, we could have a long wait to get a measurement. So they are underdamped and oscillate.

It would be possible to sense the movement of the platform and so use active damping to suppress oscillations – similar to the way our muscles control movement, as described in post 18.9. But this would be a more complicate and expensive device.

Related posts

24.6 Damped simple harmonic oscillator
24.5 Linear second-order systems
18.9 Damping and muscles

Appendix 1

Derivation of equation 6

From equations 2, 3 and 8, when p² > 1 (M²/4km > 1), we can write the solutions of our quadratic equation as

x = hexp(-pt + qt) = hexp(-pt)exp(qt)   and   x = hexp(-pt – qt) = hexp(-pt)exp(-qt).

These are two possible mathematical solutions for equation 1. But, since equation 1 is a differential equation, any linear combination of these solutions is also a solution, as explained in post 19.10. We are seeking a linear combination that has x = h when t = 0, in order to provide a solution that is a mathematical description of the physical system. Let’s explore the combination

x= (h/2)exp(-pt)exp(qt) + (h/2)exp(-pt)exp(-qt) = hexp(-pt)[exp(qt) + exp(-qt)]/2.

According to post 22.8, this result can be written as

x= hexp(-pt)cosh(qt)                    (9).

Appendix 3 shows that, from this definition, cosh(θ) = 1 when θ = 0. Therefore, equation 9 has the property that x = h when t = 0; it then starts to decrease as shown in the graph in the main text so that equation 9 describes a description of the physical system.

Appendix 2

Derivation of equation 7

From equations 2, 3 and 8, when p² < 1 (M²/4km < 1), we can write the solutions of our quadratic equation as

x = hexp(-pt + iqt) = hexp(-pt )exp(iqt)   and   x = hexp(-pt – iqt) = hexp(-pt)exp(-iqt)

where i is the square root of -1. Once again, we have two possible mathematical solutions for equation 1. And, since equation 1 is a differential equation, any linear combination of these solutions is also a solution. We are still seeking a linear combination that has x = h when t = 0 and then starts to decrease as t increases, to provide a mathematical description of the physical system. Now let’s explore the combination

x= (h/2)exp(-pt)exp(iqt) + (h/2)exp(-pt)exp(-iqt) = hexp(-pt)[exp(iqt) + exp(-iqt)]/2.

According to post 22.8 (appendix 2), this result can be written as

x= hexp(-pt)cos(qt)                    (10).

And, according to post 16.50, cos(θ) = 1 when θ = 0 and then starts to decrease as θ increases. Then, equation 10 also has the property that x = h when t = 0 and then starts to decrease as shown in the graph in the main text; so equation 10 describes a description of the physical system.

Appendix 3

To show that coshθ = 1 when θ = 0

According to post 22.8

coshθ = [exp(θ) + exp(-θ)]/2

so that, when θ = 0,

cosh(0)= [exp(0) + exp(0)]/2 = (1 + 1)/2 = 1.