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Write an expression for a transverse harmonic wave that has a wavelength of 2.5 m and propagates to the right with a speed of 13.8 m/s . The amplitude of the wave is 0.14 m , and its displacement at t=0 and x=0 is 0.14 m .

Respuesta :

Answer:

[tex]y = 0.14 Cos\left ( 2.512x-34.66t \right )[/tex]

Explanation:

wavelength, λ = 2.5 m

speed, v = 13.8 m/s

Amplitude, A = 0.14 m

The general equation of the transverse harmonic wave which is travelling right is given by

[tex]y = A Sin\left ( \frac{2\pi }{\lambda } (x - vt)+\phi \right  )[/tex]

where, Ф is phase

At t = 0, x = 0 , y = 0.14 m

0.14 = 0.14 Sin Ф

Ф = π/2

So, the equation is

[tex]y = 0.14Sin\left ( \frac{2\pi }{2.5 } (x - 13.8t)+\frac{\pi }{2} \right  )[/tex]

[tex]y = 0.14 Cos\left ( 2.512x-34.66t \right )[/tex]

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