Question

the equation of transverse travelling wave is given by y = 0.05 sin2π(0.4x-5t) where x and y are in metres and t in seconds .calculate amplitude, wavelength and frequency time period of the wave.​

Answer 1

Answer:

given in the pic above....plzz mark it as the brainliest answer...

Answer 2

The wavelength and frequency time period of the wave are 2.5 m, 5Hz and 0.2 s respectively.

Explanation:

The equation in general wave form is given as

$Y=A\sin(kx-\omega t)$

Here A is amplitude, k is wave number, $\omega$ is the angular  frequency.

The equation of transverse travelling wave is given by y = 0.05 sin2π(0.4x-5t)

On comparing,

amplitude, A =0.05 m.

Wavelength, $\lambda = \frac{2\pi}{0.8\pi}=2.5 m$

Frequency, $f=\frac{\omega}{2\pi} =\frac{10\pi}{2\pi}=5Hz$

Time period,$T=\frac{1}{f} =\frac{1}{5}=0.2s$

Thus, wavelength and frequency time period of the wave are 2.5 m, 5Hz and 0.2 s respectively.

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