Waves

JEE Main level, numerical answer, medium

22 of 142

Question 22 of 142

A transverse wave on a string is described by y=3sin⁡(36t+0.018x+π/4)y = 3\sin(36t + 0.018x + \pi/4), where x,yx, y are in cm and tt in seconds. The least distance between the two successive crests in the wave is _____ cm. (Nearest integer)
(π=3.14\pi = 3.14)

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Answer: 349

The standard equation of a transverse wave is given by y=Asin⁡(ωt+kx+ϕ)y = A\sin(\omega t + kx + \phi)

Comparing this with the given equation y=3sin⁡(36t+0.018x+π/4)y = 3\sin(36t + 0.018x + \pi/4), we get the wave number k=0.018 cm−1k = 0.018 \text{ cm}^{-1}.

The least distance between two successive crests is the wavelength λ\lambda of the wave.

We know that k=2πλk = \dfrac{2\pi}{\lambda}

λ=2πk=2×3.140.018\lambda = \dfrac{2\pi}{k} = \dfrac{2 \times 3.14}{0.018}

λ=6.280.018=628018=348.88 cm\lambda = \dfrac{6.28}{0.018} = \dfrac{6280}{18} = 348.88 \text{ cm}

Rounding to the nearest integer, we get 349 cm349 \text{ cm}.

Answer: 349349

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