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The mass-spring system – Part 3

Classical mechanics

The mass-spring system – Part 3

Here we seek to solve the equations of a mass-spring system in order to figure out the positions of the mass at different times during its harmonic motion.

In part 2, we obtained an approximate equation that tells us the force the mass experiences due to the spring, at its different positions during its motion. We now use Newton’s famous equation F=ma to find the mass’ position at different times during its travel.

What Newton says about the mass-spring

After an approximation of small displacements, we had obtained the force equation for the mass-spring as,

(1)   \begin{equation*} F(x)=- k x \end{equation*}

Using Newton’s equation for force, F=ma, we may write the above equation as,

(2)   \begin{equation*} ma=- k x \end{equation*}

We now write the acceleration as a second derivative of position. This follows from the very definition of acceleration as being the rate of change of velocity, which in turn is a rate of change of position – equivalently, the acceleration is the rate of change of the rate of change of the position! Mathematically we write this as the second derivative of position with respect to time – \frac{d^2 x}{dt^2}, so equation (2) now reads,

    \begin{align*} m\frac{d^2 x}{dt^2}=- k x \end{align*}


    \begin{align*} \frac{d^2 x}{dt^2}=- \frac{k}{m} x \end{align*}

Since \frac{k}{m} is anyway a constant for the given mass-spring system (we have the same mass of the mass, m, and the same spring of springiness k, all through the analysis), we denote it by \omega^2,

(3)   \begin{equation*} \frac{d^2 x}{dt^2}=- \omega^2 x \end{equation*}

There’s a reason we chose \omega^2=\frac{k}{m}, in particular. It has been found experimentally that the quantity \sqrt{\frac{k}{m}} is simply the frequency with which the mass oscillates about its equilibrium point. So, \omega=\sqrt{\frac{k}{m}} is called the “angular frequency”of the mass-spring system.

Now Eq. (3) is a “second order ordinary differential equation”. Now, those are some fancy adjectives that simply tell you that the equation contains terms that are second derivatives of a function (here x(t)), the derivatives being the ordinary ones (in contrast with “partial derivatives”). And so, solving this equation will give us the function x(t) – the position of the mass at different times during its simple harmonic motion – just what we have set out to find.

Finding x(t)

There are a number ways one can solve Eq. (3). Most methods require more than elementary knowledge of mathematics. To keep the math minimum, we choose the simplest of all methods – solving by inspection. In this method all you need to do is gaze at the equation and try to guess a possible solution!

Eq. (3) reveals an important property of the function x(t) – that, differentiating x(t) twice (\frac{d^2 x}{dt^2}) will give us back the original function (x(t)) with a negative sign (and a constant multiplied: \omega^2, here). The simplest function with such a property that must first happen to you must be the sine or the cosine. Recall that differentiating sine once gives cosine, and one more differentiation gives back a negative sine (\frac{d^2 \sin(t)}{dt^2}=\frac{d}{dt}\frac{d}{dt}(\sin(t))= \frac{d}{dt}\cos(t)=-\sin(t)). And the same explanation follows for the cosine function too. Therefore, \sin(t) and \cos(t) fix the required property of the function x(t) that you get back x(t) after double differentiation (with a negative sign). But we’ve left out the constant \omega^2. Differentiating x(t) twice must give us -\omega^2 x(t) and not just - x(t). To fix this we consider the sine (or cosine) function with an “argument” \omega t, i.e., x(t)=\sin(\omega t) (or equivalently x(t)=\cos(\omega t). Also, to make the solution more general, we may consider a constant multiplied with \sin(\omega t), so that x(t)=A\sin(\omega t). Now, if you doubt any of the points explained here, all you ought to do is replace x(t) in Eq. (3) by A\sin(\omega t) (or A\cos(\omega t)), and see that the equation is satisfied.

Now, the final step to obtain the most general solution to the differential equation (3) is to realize that the equation is “linear” – it does not contain terms that have higher powers (squared, cubed, etc) of the function x(t). For example, \frac{d^2 x}{dt^2}=- \omega^2 x^2 or (\frac{d^2 x}{dt^2})^2=- \omega^2 x, wouldn’t be linear.

One of the properties of linear equations is that if x_1 and x_2 are two solutions of the equation, x_1+x_2 is also a solution. In fact, any “linear combination” of the solutions is also a solution. In our case, we may interpret this as

(4)   \begin{equation*} x(t)=A\sin(\omega t)+ B\cos(\omega t) \end{equation*}

being a solution of the differential equation (3). (Do check!)

As for the constants A and B, they can be found by knowing the position at which the mass began its simple harmonic motion (the distance from the equilibrium position through which you pulled (or pushed) the block and released), and the speed with which the block began the motion. This information is called the initial conditions and can be calculated by looking at the solution at the beginning of timet=0.

And we find (from Eq. (4)) that x(0)=B (because \sin(0)=0 and \cos(0)=1). So, B is simply the initial position of the mass, x_0.

To find A, just differentiate Eq. (4) once and set t=0 again,

(5)   \begin{equation*} \frac{dx}{dt}=v(t)=A\omega\cos(\omega t)-B\omega\sin(\omega t) \end{equation*}

    \begin{align*} v(0)=A\omega \end{align*}

So, A=\frac{v(0)}{\omega} or A=\frac{v_0}{\omega}. If you know where you released the mass (x_0) and with what initial speed (v_0), you know A and B. Also note that Eq. (5) tells you the velocity of the mass at different times during the motion, as Eq. (4) tells you the position.

To make sure you aren’t lost in all these technical details, the very purpose of our endeavor, we repeat – in order to explain the motion of the mass-spring system quantitatively, we firstly found out an equation empirically, for the force of the spring on the mass; approximated the equation to make it look simple, by considering small displacements from the equilibrium position; and solved the resulting differential equation to obtain the position of the mass at different times – x(t).

So, once you’ve pulled (or pushed) the mass away from its equilibrium point and released it, with the knowledge of where you released the mass (x_0), and the initial speed resulting due to your deliberate push (or pull) while releasing, (v_0), and of course the minimum knowledge of the set up – the mass of the mass (m) and the springiness of the spring (k), you can safely bet on the position of the mass at different times (x(t)) during its motion.

Cover image by Rachel Knickmeyer

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Roshan Sawhil is a Physics postgraduate who rejoices both doing and explaining Physics. He also finds doing Philosophy as a leisure activity quite interesting. You can find and connect with him on Facebook and Twitter.

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