I was told in my class that simple harmonic motion (SHM) describes the motion of the projection onto a straight line of the motion of a particle undergoing uniform circular motion (i.e. with constant angular velocity about the circle's center).

As we know, a spring and pendulum executes SHM but their particles are not moving in circular direction. So how come they have angular frequency? $y$ in the equation describes the position on the projection of particle but as they are not moving in circular what is $y$ describing?

We do we say that a circular motion is a periodic motion but not oscillatory. But we are also saying that it's projection is in SHM. If the original particle is not in SHM, meaning it's not satisfying the condition of SHM, how can it's projection meet those conditions (when in reality that projection is not even there)? What I am trying to say that let's just consider one condition for SHM that is force should be proportional to (-) the position of the particle. The original particle doesn't satisfy that then how come its projection experience such force when the force is not even there.

Also I can't really understand if a spring and pendulum same equation of SHM how is possible that they move in different direction and still have same equation (spring moves linearly while pendulum moves parabolically).


2 Answers 2


The forces that cause that motion in each case that you described are differentbut, their mathematic description turns out to be similar, which lead to the analagous equations of motion and similar motion.

In the case of the spring, you have force that is proportional to the deformation, but in the opposite direction; that is, $F = -kx$ (where $x$ is the deviation from the point of equilibrium, that is you get an equation of motion $$ \frac{d^2x}{dt^2} = -\frac{k}{m} x$$

In the case of the pendulum, the force is the gravity force , although only a part of it propels the motion; the derived EOM is $$ \frac{d^2\alpha}{dt^2} = -\frac{mgl}{I}\sin\alpha$$ For small deviations from the balance point: $$ \frac{d^2\alpha}{dt^2} \approx -\frac{mgl}{I} \alpha$$

For the circular motion, you have constant force, but it's direction changes. You have $$ (F_x, F_y) = \vec{F} = F \frac{\vec{R}}{R} = \frac{F}{R} (x,y)$$ that leads to the following EOMs for $x$ and $y$ separately: $$ \frac{d^2x}{dt^2} = -\frac{F}{mR} x$$ $$ \frac{d^2y}{dt^2} = -\frac{F}{mR} y$$ In all the cases the force has different origin, but the EOMs are similar, all of them can be written as $$ \frac{d^2x}{dt^2} = -\omega^2 x$$ with $\omega^2 = \frac{k}{m}$ in the first case, $\frac{mgl}{I}$ in the second case and $\frac{F}{mR}$ in the third case. At this point however, the origin of the parametr $\omega^2$ becomes mostly irrelevant - the fact that EOMs look the same makes their solutions, i.e. the motions, look simialr as well.

The fact that harmonic motion is so often seen is that a lot of forces are potential forces, that is you have $F(x) = -V'(x)$, where $V(x)$ is some potential. If you know that $x_0$ is the minimum of this potential then you have $V'(x_0)=0$ and usually $V''(x_0)>0$ (this is not guaranteed but it's a generic case) and for $x\approx x_0$: $$ V(x) \approx V(x_0) + \frac12 V''(x_0) (x-x_0)^2 $$ $$ F(x) \approx - V''(x_0) (x-x_0)$$ which generates harmonic motion.


When we apply Newton's second law to a body acted on by a force directed towards a fixed point and proportional to the displacement, x, from that point, we get the famous equation $$\frac{d^2 x}{dt^2}=-\omega^2 x\ \ \ \ \ \ \ \ \ \text{in which}\ \ \ \ \ \ \ \ \ \omega^2=\frac{\text{force per unit displacement}}{\text{mass}}.$$ The functions that fit this equation are the sinusoids $$x=\sin \omega t\ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ x=\cos \omega t\ \ \ \ \ \ \ \text{or any linear combination of these.}$$ But how are the sinusoids defined? When first introduced to sines and cosines, students are told that they are ratios of side lengths in right angled triangles. Yet the textbooks draw oscillatory graphs of $x=\sin \omega t$ and $x=\cos \omega t$ in which the angle, $\omega t,$ isn't restricted to the range $0\leq \omega t \leq \pi/2$ of possible angles in a right angled triangle.

The apparent contradiction is cleared up by realising that more general definitions are needed for $x=\sin \omega t$ and $x=\cos \omega t$ for these functions to be defined for all values of $\omega t.$ The definitions are that these functions are the projections on to a diameter of the radius of a unit circle at angle $\omega t$ to the position of the radius when $\omega t=0.$

So the movement of a point around a unit circle is not in real space, but in an abstract mathematical space. But having used the unit circle to define the sine and cosine functions, the mathematics of differentiation shows that these functions fit the Newton's second law equation, and therefore represent the variation of $x$ in real space.

[Confusion sometimes arises because students often meet graphs of sine and cosine over an unrestricted range of arguments in Physics, when their previous encounters are in the context of right angled triangles. It is therefore their Physics teachers who introduce them to the circle diagram and projections on to a diameter. Students are then tempted to think of the latter as a Physics thing, rather than a mathematical one essential to understanding the meaning of sines and cosines.]

Now to consider the difficulty raised in your last paragraph. First note that a point on a pendulum moves in a circular arc, not a parabolic arc. The motion of a pendulum is not exactly shm, but for small amplitudes is approximately shm. There are several ways of showing this. One is to consider the horizontal component of the bob's motion, so we are dealing, effectively with straight line motion. Another method is to consider motion along the arc, and to let $x$ represent displacement along the arc from the equilibrium position. Even with this meaning of $x$ we can establish an approximate equation of the form $\frac{d^2 x}{dt^2}=-\omega^2 x.$

  • $\begingroup$ Can you explain my question in 3rd para ? $\endgroup$
    – user232991
    Commented Jun 15, 2019 at 13:58
  • $\begingroup$ are you trying me to tell that the particle is actually just moving in the line of projection.But while we are deriving the equation it we are just say the opposite i.e we are describing the motion of projection of particle moving in a circle. $\endgroup$
    – user232991
    Commented Jun 15, 2019 at 14:05
  • $\begingroup$ "But while we are deriving the equation it we are just say the opposite i.e we are describing the motion of projection of particle moving in a circle." Deriving which equation? But the general gist of my answer is that you should, at least temporarily, separate the motion of the physical body for which $d^2x/dt^2=-\omega^2 x$ from the motion of the point that is the projection on to a diameter of a point going round a circle with angular velocity $\omega$. $\endgroup$ Commented Jun 15, 2019 at 14:14
  • $\begingroup$ this equation -๐‘ฅ=sin๐œ”๐‘ก or ๐‘ฅ= cos๐œ”๐‘ก $\endgroup$
    – user232991
    Commented Jun 15, 2019 at 14:25
  • $\begingroup$ These equations are the solutions that fit $๐‘‘^2๐‘ฅ/๐‘‘๐‘ก^2=โˆ’๐œ”^2๐‘ฅ.$.If this equation comes from applying N2L to a particle experiencing a force proportional to its displacement along a straight line from a fixed point, and directed towards that point, then the particle is moving in THAT straight line. All the stuff about circular motion and projections is essentially the ๐‘‘๐‘’๐‘“๐‘–๐‘›๐‘–๐‘ก๐‘–๐‘œ๐‘› of the sine and cosine of an angle that increases at a constant rate, $\omega,$ with time. $\endgroup$ Commented Jun 15, 2019 at 17:02

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