Two SHMs of same frequency but different amplitudes — resultant

Question

Two simple harmonic motions (SHMs) have the same frequency $\omega$ but different amplitudes $A_1$ and $A_2$ , and a phase difference $\phi$ between them. Find the amplitude and phase of the resultant SHM.

Solution — Step by Step

Let the two SHMs be:

x_1 = A_1 \sin(\omega t)

x_2 = A_2 \sin(\omega t + \phi)

The resultant displacement is:

x = x_1 + x_2 = A_1 \sin(\omega t) + A_2 \sin(\omega t + \phi)

x_2 = A_2(\sin\omega t \cos\phi + \cos\omega t \sin\phi)

So:

x = (A_1 + A_2\cos\phi)\sin\omega t + (A_2\sin\phi)\cos\omega t

Let $R\cos\delta = A_1 + A_2\cos\phi$ and $R\sin\delta = A_2\sin\phi$ .

Then: $x = R(\cos\delta \sin\omega t + \sin\delta \cos\omega t) = R\sin(\omega t + \delta)$

Squaring and adding the two equations:

R^2 = (A_1 + A_2\cos\phi)^2 + (A_2\sin\phi)^2

= A_1^2 + 2A_1 A_2\cos\phi + A_2^2\cos^2\phi + A_2^2\sin^2\phi

= A_1^2 + 2A_1 A_2\cos\phi + A_2^2

\boxed{R = \sqrt{A_1^2 + A_2^2 + 2A_1 A_2\cos\phi}}

\tan\delta = \frac{R\sin\delta}{R\cos\delta} = \frac{A_2\sin\phi}{A_1 + A_2\cos\phi}

\boxed{\delta = \tan^{-1}\!\left(\frac{A_2\sin\phi}{A_1 + A_2\cos\phi}\right)}

The resultant SHM is: $x = R\sin(\omega t + \delta)$ with frequency $\omega$ (same as both components).

Why This Works

This is analogous to vector addition — superimposing two sinusoids of the same frequency gives another sinusoid of the same frequency (but different amplitude and phase). This is a fundamental property of linear systems.

Special cases:

$\phi = 0$ (same phase): $R = A_1 + A_2$ (maximum, constructive interference)
$\phi = \pi$ (opposite phase): $R = |A_1 - A_2|$ (minimum, destructive interference)
$\phi = \pi/2$ (90° apart): $R = \sqrt{A_1^2 + A_2^2}$ (like perpendicular vectors)

Common Mistake

Students often try to add amplitudes directly: $R = A_1 + A_2$ for all cases. This is only valid when $\phi = 0$ (same phase). In general, the resultant amplitude depends on the phase difference $\phi$ . The correct formula is $R = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos\phi}$ — this is essentially the cosine rule for triangle with sides $A_1$ and $A_2$ and included angle $\pi - \phi$ .

The formula $R = \sqrt{A_1^2 + A_2^2 + 2A_1 A_2\cos\phi}$ is identical to the parallelogram law of vector addition. Think of $A_1$ and $A_2$ as two vectors with an angle $\phi$ between them. The resultant is their vector sum. This analogy makes the formula easy to recall and understand geometrically.

Question

Solution — Step by Step

Why This Works

Common Mistake

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