https://thinking-about-science.com/2024/09/08/24-8-passive-damping-bathroom-scales/ Skip to content Thinking about Science with David Hukins Understanding science NOT learning "facts" Menu * INDEX * Reading this blog * Who am I? * What's this blog for? 24.8 Passive damping - bathroom scales September 8, 2024 David Hukins [fig1] 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 [fig2] 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(d^2x/dt^2) 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 d^2x/dt^2 = Aa^2e^at. Then equation 1 becomes A(k + Ma + ma^2)e^at = 0. Since e^at cannot equal zero, the possible solutions to this equation are A = 0 and (k + Ma + ma^2) = 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 [fig4] 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 x[0] = h. (4) We are now going to consider three possible cases for the behaviour of our scales depending on the relative values of M^2 and 4km. Case 1 M^2/4km = 1 In this case, the expression under the square root sign of equation 3 is zero. Then, from equations 2, 3 and 4, [fig5] Here exp(y) is another way of writing e^y. This is called critical damping. Case 2 M^2/4km > 1 Now the expression under the square root sign of equation 3 is a real number. In appendix 1, I show that now [fig6] Here cosh(y) is the hyperbolic cosine of y, defined in post 22.8. This is called overdamping. Case 3 M^2/4km < 1 Now the expression under the square root sign of equation 3 is an imaginary number. In appendix 2, I show that now [fig7] This is called underdamping. Comparison of the three cases To compare the three cases, I define [fig8] The reason is to reduce the number of symbols in our equations. The table below then summarises all the results so far. [fig9] [fig3] 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 p ^2 = 1 (that is M^2/4km = 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 p^2 > 1 (M^2/4km > 1), we can write the solutions of our quadratic equation as x = hexp(-pt + qt) = hexp(-pt)exp(qt) and x = hexp(-pt - qt) = h exp(-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(th) = 1 when th = 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 p^2 < 1 (M^2/4km < 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(th) = 1 when th = 0 and then starts to decrease as th 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 coshth = 1 when th = 0 According to post 22.8 coshth = [exp(th) + exp(-th)]/2 so that, when th = 0, cosh(0)= [exp(0) + exp(0)]/2 = (1 + 1)/2 = 1. Share this: * Twitter * Facebook * Like Loading... Share this: * Twitter * Facebook * Like Loading... Post navigation Previous Post 24.7 Driven oscillator One thought on "24.8 Passive damping - bathroom scales" 1. Pingback: Passive Damping - Yu Shi Cheng - Pian Zhi De Ma Nong Leave a comment Cancel reply [ ] [ ] [ ] [ ] [ ] [ ] [ ] D[ ] Recent Posts * 24.8 Passive damping - bathroom scales * 24.7 Driven oscillator * 24.6 Damped simple harmonic oscillator * 24.5 Linear second-order systems * 24.4 Logistic difference equation: self-controlled growth and chaos Recent Comments Passive Damping - Yu Shi ... on 24.8 Passive damping - bathroo... [4f7ba] David Hukins on 20.10 Toughness [53592] Peter Purslow on 20.10 Toughness [d1742] strangeset on 19.8 Wave energy [d1742] strangeset on 16.1 Drug safety testing Archives * September 2024 * August 2024 * June 2024 * May 2024 * April 2024 * March 2024 * June 2023 * May 2023 * April 2023 * March 2023 * February 2023 * January 2023 * December 2022 * November 2022 * October 2022 * September 2022 * August 2022 * July 2022 * June 2022 * May 2022 * April 2022 * March 2022 * February 2022 * November 2021 * October 2021 * September 2021 * August 2021 * July 2021 * June 2021 * May 2021 * April 2021 * March 2021 * February 2021 * January 2021 * December 2020 * November 2020 * October 2020 * September 2020 * August 2020 * July 2020 * June 2020 * May 2020 * April 2020 * March 2020 * February 2020 * January 2020 * December 2019 * November 2019 * October 2019 * September 2019 * August 2019 * July 2019 * June 2019 * May 2019 * April 2019 * March 2019 * February 2019 * January 2019 * December 2018 * November 2018 * October 2018 * September 2018 * August 2018 * July 2018 * June 2018 * May 2018 * April 2018 * March 2018 * February 2018 * January 2018 * December 2017 * November 2017 * October 2017 * September 2017 * August 2017 * July 2017 * June 2017 * May 2017 * April 2017 * March 2017 * February 2017 * January 2017 * December 2016 * November 2016 * October 2016 * September 2016 * August 2016 * July 2016 * June 2016 * May 2016 * April 2016 * March 2016 * February 2016 * January 2016 Follow Blog via Email Enter your email address to follow this blog and receive notifications of new posts by email. Email Address: [ ] Follow Create a free website or blog at WordPress.com. [Close and accept] Privacy & Cookies: This site uses cookies. By continuing to use this website, you agree to their use. To find out more, including how to control cookies, see here: Cookie Policy * Comment * Reblog * Subscribe Subscribed + [wpcom-] Thinking about Science with David Hukins Join 38 other subscribers [ ] Sign me up + Already have a WordPress.com account? Log in now. * + [wpcom-] Thinking about Science with David Hukins + Customize + Subscribe Subscribed + Sign up + Log in + Copy shortlink + Report this content + View post in Reader + Manage subscriptions + Collapse this bar Loading Comments... Write a Comment... [ ] Email (Required) [ ] Name (Required) [ ] Website [ ] [Post Comment] %d %d [b]