Question

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 .

Answer 1

Answer:

$y = 0.14 Cos\left ( 2.512x-34.66t \right )$

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

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

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

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

$y = 0.14 Cos\left ( 2.512x-34.66t \right )$