Trigonometric Identities: Real-World Scenarios (6)

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Question

A pendulum’s projected horizontal motion follows x(t)=5sin(2t)+12cos(2t)x(t) = 5\sin(2t) + 12\cos(2t). Express this as a single sinusoid Rsin(2t+ϕ)R\sin(2t + \phi) and find the maximum displacement.

Solution — Step by Step

asinθ+bcosθ=Rsin(θ+ϕ)a\sin\theta + b\cos\theta = R\sin(\theta + \phi)

where R=a2+b2R = \sqrt{a^2 + b^2} and tanϕ=b/a\tan\phi = b/a.

For our problem: a=5a = 5, b=12b = 12, θ=2t\theta = 2t.

R=52+122=25+144=169=13R = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13

The maximum displacement is the amplitude R=13R = 13.

tanϕ=12/5\tan\phi = 12/5

ϕ=tan1(12/5)67.38°\phi = \tan^{-1}(12/5) \approx 67.38° or about 1.1761.176 radians.

x(t)=13sin(2t+tan1(12/5))x(t) = 13 \sin(2t + \tan^{-1}(12/5))

Maximum displacement: 1313 units.

Why This Works

The harmonic addition formula uses the geometric fact that any linear combination of sin\sin and cos\cos at the same frequency is itself a sinusoid at that frequency, just shifted in phase.

Geometrically: asin+bcosa\sin + b\cos is the dot product of (a,b)(a, b) with (sinθ,cosθ)(\sin\theta, \cos\theta), which traces a circle of radius 1. The maximum dot product is a2+b2\sqrt{a^2 + b^2}, achieved when (a,b)(a, b) aligns with the unit vector.

Speed shortcut: For asin+bcosa\sin + b\cos, the amplitude is a2+b2\sqrt{a^2 + b^2} — write this immediately. Saves time when the phase is not asked.

Alternative Method — Use Cosine Addition

We could also write asinθ+bcosθ=Rcos(θα)a\sin\theta + b\cos\theta = R\cos(\theta - \alpha) where tanα=a/b\tan\alpha = a/b.

Both forms give the same amplitude a2+b2\sqrt{a^2 + b^2} — only the phase representation changes. Use whichever the question asks for.

Common Mistake

Students often compute R=a+bR = a + b instead of a2+b2\sqrt{a^2 + b^2}, getting 5+12=175 + 12 = 17 instead of 1313. The amplitudes don’t add linearly — they add in quadrature (Pythagoras).

Another classic: confusing tanϕ=b/a\tan\phi = b/a with tanϕ=a/b\tan\phi = a/b. Memorise one form clearly: for sin(θ+ϕ)\sin(\theta + \phi) form, tanϕ=b/a\tan\phi = b/a where bb is the cosine coefficient and aa is the sine coefficient.

JEE Main and CBSE boards both ask harmonic-combination problems regularly. The 5-12-13 and 3-4-5 triangles are favourite test setups because they give clean Pythagorean amplitudes. NEET physics also uses this in superposition of two perpendicular SHMs at the same frequency.

This formula has direct physics applications: combining two waves at the same frequency, AC circuit phasor sums, and resolving force components.

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