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NCERT Exemplar · Q19

Q.The displacement of an elastic wave is given by the function y=3sin⁡ωt+4cos⁡ωty = 3\sin\omega t + 4\cos\omega t. where yy is in cm and tt is in second. Calculate the resultant amplitude.

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Two perpendicular harmonic components of the same frequency combine vectorially; the resultant amplitude is found by treating the sine and cosine coefficients as perpendicular vectors. Resultant amplitude = 5 cm.

Why This Works: Superposition of Perpendicular Components

When two harmonic oscillations of the same angular frequency ω\omega are added, they interfere to produce a single harmonic motion at that frequency. The key insight is that sin⁡ωt\sin\omega t and cos⁡ωt\cos\omega t are perpendicular in phase space—they differ by 90°90°. Just as perpendicular vectors add by the Pythagorean theorem, so do these components.

The general principle: any linear combination Asin⁡ωt+Bcos⁡ωtA\sin\omega t + B\cos\omega t can be rewritten as a single sinusoid Rsin⁡(ωt+ϕ)R\sin(\omega t + \phi), where the resultant amplitude RR is the vector sum of the two perpendicular contributions.

Step-by-Step Solution

1. Identify the two harmonic components

The displacement is:

y=3sin⁡ωt+4cos⁡ωty = 3\sin\omega t + 4\cos\omega t

Here the coefficient of sin⁡ωt\sin\omega t is A=3A = 3 cm and the coefficient of cos⁡ωt\cos\omega t is B=4B = 4 cm.

2. Recognize the phase relationship

Since cos⁡ωt=sin⁡(ωt+90°)\cos\omega t = \sin(\omega t + 90°), the two terms oscillate at the same frequency but are 90°90° out of phase. This perpendicularity means we cannot simply add the amplitudes arithmetically; instead, they combine as perpendicular vectors.

3. Apply the vector addition formula

The resultant amplitude RR of two perpendicular oscillations with amplitudes AA and BB is:

R=A2+B2R = \sqrt{A^2 + B^2}

Substituting our values:

R=32+42=9+16=25=5 cmR = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ cm} …

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