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NCERT Exemplar · Q6

Q.The displacement of a particle varies with time according to the relation y=asin⁡ωt+bcos⁡ωty = a\sin\omega t + b\cos\omega t.

(a) The motion is oscillatory but not S.H.M.
(b) The motion is S.H.M. with amplitude a+ba + b.
(c) The motion is S.H.M. with amplitude a2+b2a^{2} + b^{2}.
(d) The motion is S.H.M. with amplitude a2+b2\sqrt{a^{2} + b^{2}}.
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A linear combination of sin⁡ωt\sin\omega t and cos⁡ωt\cos\omega t with the same angular frequency ω\omega is equivalent to a single sine (or cosine) function with a phase shift — so the motion is simple harmonic with amplitude a2+b2\sqrt{a^{2}+b^{2}}.

The key insight is that any sum of two sinusoidal functions of the same frequency is itself a sinusoid of that frequency. That is the hallmark of simple harmonic motion: a restoring force proportional to displacement leads to a sinusoidal solution, and here the displacement is exactly that — a single sine wave, just written in a different form.

Let’s see why.

  1. Recognise the form. The given equation is

y=asin⁡ωt+bcos⁡ωt.y = a\sin\omega t + b\cos\omega t.

Both terms oscillate with the same angular frequency ω\omega. This is not two independent frequencies; it is one frequency expressed as a sum of a sine and a cosine.

  1. Combine into a single sinusoid. Any expression of the form Asin⁡θ+Bcos⁡θA\sin\theta + B\cos\theta can be rewritten as

Rsin⁡(θ+ϕ)orRcos⁡(θ−ϕ),R\sin(\theta + \phi) \quad \text{or} \quad R\cos(\theta - \phi),

where R=A2+B2R = \sqrt{A^{2}+B^{2}} and ϕ\phi is a phase angle such that tan⁡ϕ=BA\tan\phi = \frac{B}{A} (for the sine form).

Here, take A=aA = a, B=bB = b, and θ=ωt\theta = \omega t. Then

y=a2+b2sin⁡(ωt+ϕ),where ϕ=tan⁡−1 ⁣(ba).y = \sqrt{a^{2}+b^{2}} \sin(\omega t + \phi), \quad \text{where } \phi = \tan^{-1}\!\left(\frac{b}{a}\right).

Tip

To derive this quickly: multiply and divide by a2+b2\sqrt{a^{2}+b^{2}}, then set cos⁡ϕ=aa2+b2\cos\phi = \frac{a}{\sqrt{a^{2}+b^{2}}} and sin⁡ϕ=ba2+b2\sin\phi = \frac{b}{\sqrt{a^{2}+b^{2}}}. The sum becomes a2+b2(sin⁡ωtcos⁡ϕ+cos⁡ωtsin⁡ϕ)=a2+b2sin⁡(ωt+ϕ)\sqrt{a^{2}+b^{2}}(\sin\omega t \cos\phi + \cos\omega t \sin\phi) = \sqrt{a^{2}+b^{2}}\sin(\omega t+\phi).

  1. Identify the motion. …

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