Wave Equation

Question

A travelling wave is described by y equals 0.02 sin of two pi times bracket 5 t minus 0.25 x in SI units, so what is the wavelength of this wave?

RankGuru · Accepted Answer

Writing the phase in the convenient shape $y = A\sin\!\left[2\pi\left(\dfrac{t}{T} - \dfrac{x}{\lambda}ight)ight]$ used in NCERT Class 11, Chapter 15 (Waves), the number multiplying $t$ is $1/T$ and the number multiplying $x$ is $1/\lambda$. Comparing this with the given $y = 0.02\sin[2\pi(5t - 0.25x)]$, we read off $1/T = 5$ Hz and $1/\lambda = 0.25$ per metre. The wavelength is therefore the reciprocal of the coefficient of $x$: $\lambda = \dfrac{1}{0.25} = 4$ m. Equivalently, expanding gives a wave number $k = 2\pi(0.25) = \pi/2$ rad/m, and $\lambda = 2\pi/k = 2\pi/(\pi/2) = 4$ m, which agrees. The option 2 m mistakenly treats the coefficient as $0.5$ rather than $0.25$. The option 0.25 m reports the coefficient itself instead of taking its reciprocal. The option 10 m confuses wavelength with a quantity built from the temporal term. As a further cross-check one can combine the two readings to find the wave speed: with $T = 1/5 = 0.2$ s and $\lambda = 4$ m, the speed is $v = \lambda/T = 4/0.2 = 20$ m/s, a perfectly ordinary value that keeps the whole description internally consistent. A plausibility check: the reciprocal of a spatial frequency of $0.25\,	ext{m}^{-1}$ has units of metres and gives a physically reasonable crest spacing of a few metres, confirming both the magnitude and the dimensions of the result.

Wave Equation

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