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

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

Step-by-step 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 ( (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 ( (2\pi )/(2.5 ) (x - 13.8t)+(\pi )/(2) \right  )`

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