Will the combination of 2 Simple Harmonic motions will be an SHM in itself? For example for simple functions such as

$$\ f(t)=\sin\omega t-\cos\omega t$$ I can use trigonometry to show that it can be expressed as $$\ f(t)=\sqrt 2\sin(\omega t-\pi/4) $$.

But what about functions given in the questions given below?

[Ref: “NCERT Class 11th (XI) Physics, Part 2”, Digital Designs; notes on p. 357 and Problem 14.4, p. 359 <link> ]

In (b) I can express the function as a combination of

$\sin\omega t$ and $\sin3\omega t$.

Each of these 2 terms can independently express an SHM but will their combination do the same?

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As an answer to part (b) and (d) ,the book says that the superposition of two SHM is always periodic but never an SHM. (I believe that this is incorrect.Maybe a typo)

Moreover, at the end of the chapter there is a note: enter image description here

I am getting pretty confused.

Can anybody tell me when the combination of 2 SHM's be an SHM/periodic/not periodic?

  • $\begingroup$ This question and the provided answers could be helpful. @Karan Singh $\endgroup$ – AlQuemist Dec 3 '15 at 9:23
  • $\begingroup$ I think the defining property of a simple harmonic motion is that it is a periodic motion with a constant amplitude and a constant phase; that is, it should be possible to describe it as $x(t) = A \, \cos(\omega t + \phi)$, where $A$, $\omega$, and $\phi$ are time-independent constants. @Karan Singh $\endgroup$ – AlQuemist Dec 3 '15 at 9:31
  • $\begingroup$ It is not $\sin 3 \omega t$ but $\sin^3 \omega t$ in the textbook question. $\endgroup$ – ja72 Dec 3 '15 at 13:28

Consider the superposition of two simple harmonic motions \begin{equation} x(t) = x_1(t) + x_2(t) = A_1 \cos \left( \omega_1 t + \phi_1 \right) + A_2 \cos \left( \omega_2 t + \phi_2 \right). \end{equation} The first motion $x_1(t)$ is periodic with period $T_1 = \frac{2\pi}{\omega_1}$ and the second motion $x_2(t)$ is periodic with period $T_2 = \frac{2\pi}{\omega_2}$. Clearly the sum of both is only periodic if $n T_1 = m T_2$ where $n$ and $m$ are positive integers (thanks to user fibonatic for pointing out the most general case). To see this, simply write \begin{equation} x\left(t+nT_1\right) = x_1\left(t+nT_1\right) + x_2\left(t+mT_2\right) = x_1(t) + x_2(t) = x(t). \end{equation}

Moreover if the period of both harmonic motions is the same $\omega_1 = \omega_2 = \omega$, we can write \begin{align} x(t) & = A_1 \left[ \cos(\omega t) \cos \phi_1 - \sin(\omega t) \sin \phi_1 \right] + A_2 \left[ \cos(\omega t) \cos \phi_2 - \sin(\omega t) \sin \phi_2 \right] \\ & = \left[ A_1 \cos \phi_1 + A_2 \cos \phi_2 \right] \cos(\omega t) - \left[ A_1 \sin(\phi_1) + A_2 \sin \phi_2 \right] \sin(\omega t) \\ & = A \cos \left( \omega t + \phi \right), \end{align} where used the sum rule $\cos\left( \alpha + \beta \right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$, and we defined \begin{align} A \cos \phi & = A_1 \cos \phi_1 + A_2 \cos \phi_2 \\ A \sin \phi & = A_1 \sin \phi_1 + A_2 \sin \phi_2 . \end{align} This can be generalized to an arbitrary sum of harmonic motions with the same period: \begin{equation} \sum_i A_i \cos \left( \omega t + \phi_i \right) = A \cos \left( \omega t + \phi \right). \end{equation} Another way to understand this, is to note that the harmonic equation is a linear differential equation; any linear combination of solutions is also a solution.


The sum of two harmonic motions with frequencies $\omega_1$ and $\omega_2$ is periodic if the ratio $\frac{\omega_1}{\omega_2}$ is a positive rational number. If the ratio is irrational, the resulting motion is not periodic.

If, moreover, the frequencies of the two harmonic motions are equal, the resulting motion is also a harmonic motion with the same frequency.

  • $\begingroup$ I would have to disagree with you, for example according to your definition $\omega_1=3$ and $\omega_2=5$ should not be periodic, but it is. Namely it is periodic if $\frac{\omega_1}{\omega_2}=\frac{n}{m}$, where $n$ and $m$ are both positive integers. $\endgroup$ – fibonatic Dec 3 '15 at 18:26
  • $\begingroup$ @fibonatic Indeed, I have changed my answer. Thanks. $\endgroup$ – Praan Dec 3 '15 at 18:54
  • $\begingroup$ @KaranSingh I have updated my answer. I made an oversight concerning the periodicity. See the comment of fibonatic. $\endgroup$ – Praan Dec 3 '15 at 19:29

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