Combining Two Oscillations

What if two restoring forces act on the same particle at once? The answer is pleasantly simple: the motions just add.

Motions add#

Suppose force F₁ alone would move a particle along $\vec r_1(t)$, and F₂ alone along $\vec r_2(t)$. Then $m\ddot{\vec r}_1 = \vec F_1$ and $m\ddot{\vec r}_2 = \vec F_2$. Adding, $m(\ddot{\vec r}_1 + \ddot{\vec r}_2) = \vec F_1 + \vec F_2$, so $\vec r = \vec r_1 + \vec r_2$ is the motion under both forces (with matching starting conditions). Find each motion separately and add the displacements.

Two SHMs along the same line#

Take equal frequencies with a phase difference δ:

Expanding the second and collecting, $x = (A_1 + A_2\cos\delta)\sin\omega t + (A_2\sin\delta)\cos\omega t$, which can be rewritten as a single sine:

The result is another SHM of the same frequency. Its amplitude depends on the phase difference:

• δ = 0 (in phase): $A = A_1 + A_2$, the largest possible.
• δ = π (out of phase): $A = |A_1 - A_2|$; equal amplitudes cancel completely and the particle stays still.
• Anything in between gives an amplitude between these limits.

The phasor (vector) method

Represent each SHM by an arrow whose length is its amplitude, drawn at an angle equal to its phase: A₁ along a reference line, A₂ at angle δ to it. Their vector sum has exactly the magnitude and angle given above. It works for any number of SHMs of the same frequency: draw the arrows head to tail.

Two SHMs at right angles#

Now let one force produce motion along x and the other along y, again at the same frequency:

The particle moves in a plane. Eliminating t gives the path:

• δ = 0: $y = (A_2/A_1)x$, a straight line along one diagonal of the rectangle: SHM of amplitude $\sqrt{A_1^2 + A_2^2}$.
• δ = π: a straight line along the other diagonal.
• δ = π/2: $x^2/A_1^2 + y^2/A_2^2 = 1$, an ellipse with axes along x and y. With A₁ = A₂ it is a circle: two equal perpendicular SHMs a quarter-cycle apart make uniform circular motion (clockwise or anticlockwise depending on which leads).