Which of the following equations represents a travelling wave?

Waves PYQ — JEE Main 2021 · 24 Feb · Shift 2 · Q53 with Solution

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Correct answer: C

General form of a travelling wave

A one-dimensional travelling wave is represented by a function of the form

y(x,t)=f(x-vt) \quad \text{or} \quad f(x+vt),

and sinusoidal travelling waves are commonly written as

y(x,t)=A\sin(kx-\omega t+\phi)

y(x,t)=A\cos(kx-\omega t+\phi).

So, for a wave to be a travelling wave, the space variable $x$ and time variable $t$ must appear together in the combination $kx\pm \omega t$ (or equivalently $x\pm vt$ ).

Check each option

Option A: $y=Ae^x\cos(\omega t-\theta)$

This is of the form

y=(\text{function of }x)\times(\text{function of }t).

Here, $x$ and $t$ do not appear in the combined form $x\pm vt$ or $kx\pm \omega t$ . So this is not a travelling wave.

Option B: $y=Ae-x2(vt+\theta)$

As written, this expression is not in a standard travelling-wave form. It does not clearly represent a function of $x\pm vt$ . Hence, this is not a travelling wave.

Option C: $y=A\sin(15x-2t)$

This is exactly of the form

y=A\sin(kx-\omega t),

with

k=15, \qquad \omega=2.

Therefore, this is a travelling wave.

Option D: $y=A\sin x\cos \omega t$

This is again a product of a function of $x$ and a function of $t$ :

y=(A\sin x)(\cos \omega t).

This represents a standing wave, not a travelling wave.

Conclusion

The only equation that represents a travelling wave is

\boxed{y=A\sin(15x-2t)}

So, the correct option is C.

Waves question

Waves question

Option A: $y=Ae^x\cos(\omega t-\theta)$

Option B: $y=Ae-x2(vt+\theta)$

Option C: $y=A\sin(15x-2t)$

Option D: $y=A\sin x\cos \omega t$

More from Waves

Waves question

Waves question

Option A: $y=Ae^x\cos(\omega t-\theta)$

Option B: $y=Ae-x2(vt+\theta)$

Option C: $y=A\sin(15x-2t)$

Option D: $y=A\sin x\cos \omega t$

More from Waves

Option A: y=Aexcos⁡(ωt−θ)y=Ae^x\cos(\omega t-\theta)y=Aexcos(ωt−θ)

Option B: y=Ae−x2(vt+θ)y=Ae-x2(vt+\theta)y=Ae−x2(vt+θ)

Option C: y=Asin⁡(15x−2t)y=A\sin(15x-2t)y=Asin(15x−2t)

Option D: y=Asin⁡xcos⁡ωty=A\sin x\cos \omega ty=Asinxcosωt

More from Waves

Option A: y=Aexcos⁡(ωt−θ)y=Ae^x\cos(\omega t-\theta)y=Aexcos(ωt−θ)

Option B: y=Ae−x2(vt+θ)y=Ae-x2(vt+\theta)y=Ae−x2(vt+θ)

Option C: y=Asin⁡(15x−2t)y=A\sin(15x-2t)y=Asin(15x−2t)

Option D: y=Asin⁡xcos⁡ωty=A\sin x\cos \omega ty=Asinxcosωt

More from Waves

Option A: $y=Ae^x\cos(\omega t-\theta)$

Option B: $y=Ae-x2(vt+\theta)$

Option C: $y=A\sin(15x-2t)$

Option D: $y=A\sin x\cos \omega t$

Option A: $y=Ae^x\cos(\omega t-\theta)$

Option B: $y=Ae-x2(vt+\theta)$

Option C: $y=A\sin(15x-2t)$

Option D: $y=A\sin x\cos \omega t$