I'm following Classical Mechanics, 5th Edition by Tom W.B. Kibble and Frank H. Berkshire. I'm following it since I'm interested in studying physics (although, am doing it at home myself).

I've worked through quite a range of chapters in the book but skipped a lot of chapter two since I was having so much trouble understanding it (although a lot of the rest was fine).

The section I'm struggling with is where they solve the harmonic oscillator equation.

Equation 2.13 ($m\ddot{x} + kx = 0$) is a linear differential equation; that is, one involving only linear terms in x and its derivatives. Such equations have the important property that their solutions satisfy the superposition principle: if $x_1(t)$ and $x_2(t)$ are solutions, then so is any linear combination $$x(t)=a_1x_1(t)+a_2x_2(t), \quad[2.15]$$ where $a_1$ and $a_2$ are constants; for, clearly, $$m\ddot{x}+kx = a_1(m\ddot{x_1}+kx_1) + a_2(m\ddot{x_2}+kx_2) = 0$$

This makes sense so far - they're just talking about the property of solutions to differential equations.

Moreover if $x_1$ and $x_2$ are linearly independent solutions, then [2.15] is the general solution. Since the equation is of second order, we could obtain its solution by integrating twice, and the general solution must therefore contain just two abritrary constants of integratio. So to find the general solution, all we have to do is find any two independent solutions $x_1(t)$ and $x_2(t)$.

Let us first consider the case where $k < 0$, so that $V(x)$ has a maximum at $x = 0$. Then, the differential equation can be written, $$\ddot{x} - p^2x = 0, \quad p = \sqrt{\frac{-k}{m}}$$

This is the part I don't understand. Where on earth did p come from? What is p? How did they actually solve this to get the general solution $x = \frac{1}{2}Ae^{pt} + \frac{1}{2}Be^{-pt}$? They then go on to turn to the case where $k > 0$ and this doesn't make sense either.

No matter how many times I read through it I don't get it. Would it be better to understand the solution with complex numbers rather than these two different scenarios for k?

Thanks for any help.

  • 2
    $\begingroup$ Bear in mind that this was written as an undergraduate text in the UK, and (from the preface) "It is designed for students with some previous acquaintance with the elementary concepts of mechanics … An essential prerequisite is a reasonable familiarity with differential and integral calculus, including partial differentiation." In other words, the intended readership would already know what simple harmonic motion was, so the book doesn't spend time deriving it carefully from first principles. First-year US university textbooks tend to assume less prior knowledge. $\endgroup$
    – alephzero
    Feb 26, 2019 at 17:28
  • $\begingroup$ ... the thing that would be new to the intended readers would be the emphasis on kinetic and potential energy (i.e. the functions $T$ and $V$) not the actual results about SHM. $\endgroup$
    – alephzero
    Feb 26, 2019 at 17:33
  • $\begingroup$ One issue to keep in mind, is that the exponent term for $e$ must be dimensionless. This means that the term $pt$ must be dimensionless. Due to this, $p$ must have a dimension of $1/t$, which will make it a frequency term. $\endgroup$ Feb 26, 2019 at 20:39
  • $\begingroup$ "No matter how many times I read through it I don't get it." - just a general hint for 'reading' textbooks: reading usually isn't enough. Copy out the equations. Try to derive it yourself, fill in the derivation not shown. You're clearly thinking along and maybe trying to do it in your head- works better on paper if you're in the habit already! $\endgroup$
    – jklebes
    Mar 3, 2021 at 8:55
  • $\begingroup$ In my answer here : Need help understanding an equation of motion for a pendulum you will find a very simple method (trick I would say) to solve the 2nd order linear differential equation of the harmonic oscillator. $\endgroup$
    – Frobenius
    Mar 3, 2021 at 14:56

2 Answers 2


If $k<0$ it is possible to define $p$ such that $k=-{{p}^{2}}$ No more that a change of notation !

Then you can easily check that ${{e}^{pt}}$ and ${{e}^{-pt}}$ are two independant solutions and the general solution is a linear combination of these two (The $1/2$ factor is not necessary, just a notation !)

  • $\begingroup$ It is a change in notation but you are missing the "m" in the expression. $\endgroup$
    – nasu
    Jul 3, 2021 at 11:57

If you start with the equation $m\ddot{x}+kx=0 $ and divide all terms by m you get $\ddot{x}+\frac{k}{m}x=0 $. Then you just label the ratio $\frac{k}{m} $ as $-p^2$. It is a standart step when dealing with this equation and the author does not show it explicitly. For the more coomon case of positive k, the ratio is usually labeled as $\omega^2$ and is called angular frequency.


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