Physics · Ch 5 — Oscillations

Composition of two S.H.M.s having same period and along the same path

5.10

Composition of two S.H.M.s having same period and along the same path

What happens if a single particle is subjected to TWO different S.H.M.s at once, both having the same period and both acting along the same straight-line path (say, the x-axis), but with different amplitudes and different initial phases? Let the two individual motions be

x_1 = A_1\sin(\omega t+\phi_1), \qquad x_2 = A_2\sin(\omega t+\phi_2)

Since both act along the same line, the resultant displacement at any instant is simply the algebraic sum, $x = x_1 + x_2$ . Expanding both sine terms using the compound-angle formula $\sin(\theta+\alpha) = \sin\theta\cos\alpha+\cos\theta\sin\alpha$ ,

x = A_1\sin\omega t\cos\phi_1 + A_1\cos\omega t\sin\phi_1 + A_2\sin\omega t\cos\phi_2 + A_2\cos\omega t\sin\phi_2

Since $A_1, A_2, \phi_1, \phi_2$ are all CONSTANTS while $\omega t$ is the variable, we can collect the coefficients of $\sin\omega t$ and of $\cos\omega t$ separately:

x = (A_1\cos\phi_1+A_2\cos\phi_2)\sin\omega t + (A_1\sin\phi_1+A_2\sin\phi_2)\cos\omega t

Now introduce two NEW constants, R and $\delta$ , defined by

R\cos\delta = A_1\cos\phi_1+A_2\cos\phi_2 \qquad \text{...(5.17)}

R\sin\delta = A_1\sin\phi_1+A_2\sin\phi_2 \qquad \text{...(5.18)}

(this is always possible for any values of the right-hand sides, since R and $\delta$ are just two new unknowns fit to the two known combinations above). Substituting Eqs. (5.17)-(5.18) into the expression for x,

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

This is remarkable: x is again a PURE S.H.M., of the SAME angular frequency $\omega$ (hence the same period) as the two individual motions, but with a NEW amplitude R and a NEW initial phase $\delta$ . In other words, superposing (adding) two S.H.M.s of the same period, along the same path, always produces a third S.H.M. of that same period.

To find R and $\delta$ explicitly: squaring and adding Eqs. (5.17) and (5.18),

R^2 = (A_1\cos\phi_1+A_2\cos\phi_2)^2+(A_1\sin\phi_1+A_2\sin\phi_2)^2 = A_1^2+A_2^2+2A_1A_2\cos(\phi_1-\phi_2) \qquad \text{...(5.19)}

(using $\cos^2+\sin^2=1$ and the compound-angle formula for $\cos(\phi_1-\phi_2)$ to simplify the cross terms). Dividing Eq. (5.18) by Eq. (5.17) instead gives

\tan\delta = \frac{A_1\sin\phi_1+A_2\sin\phi_2}{A_1\cos\phi_1+A_2\cos\phi_2} \qquad \text{...(5.20)}

Three special cases of the phase difference $(\phi_1-\phi_2)$ between the two component S.H.M.s are worth knowing by heart:

IN PHASE, $(\phi_1-\phi_2) = 0°$ , so $\cos(\phi_1-\phi_2)=1$ : Eq. (5.19) gives $R = \sqrt{A_1^2+A_2^2+2A_1A_2} = A_1+A_2$ -- the resultant amplitude is simply the SUM of the two, the largest possible resultant. If additionally $A_1=A_2=A$ , then $R = 2A$ .
90 DEGREES OUT OF PHASE, $(\phi_1-\phi_2)=90°$ , so $\cos(\phi_1-\phi_2)=0$ : Eq. (5.19) gives $R = \sqrt{A_1^2+A_2^2}$ (a "Pythagorean" combination). If $A_1=A_2=A$ , then $R = \sqrt2\,A$ .
180 DEGREES OUT OF PHASE, $(\phi_1-\phi_2)=180°$ , so $\cos(\phi_1-\phi_2)=-1$ : Eq. (5.19) gives $R = \sqrt{A_1^2+A_2^2-2A_1A_2} = |A_1-A_2|$ -- the smallest possible resultant. If $A_1=A_2=A$ , then $R = 0$ : the two motions cancel EXACTLY, and the particle does not move at all. …

Misc Activity.5.1Classroom demonstration: energy transfer between pendula of equal length via a shared string

Worked out. A horizontal string is tied tautly between two vertical supports, and three pendula are hung from it: two of them, A and B, of EQUAL length, and a third, C, of some different (but not very different) length. Setting pendula A and B oscillating together in a plane perpendicular to the horizontal string, it is observed that pendulum C ALSO begins oscillating in the same plane, with the same period as A and B. The activity demonstrates the physical content of section 5.10's mathematics: imposing two S.H.M.s of the same period (from A and B) on the shared string, whose resultant energy transfers along the string into the third pendulum C, setting it oscillating too. The text notes this same set-up can be used to verify the three special cases (in-phase, 90-degree, and 180-degree phase differences) of the resultant-amplitude formula by suitably varying how A and …

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