Respuesta :
Answer:
A) The wave equation is given as
[tex]y(x,t) = A\cos(kx + \omega t)=(3.3\times 10^{-2})\cos(0.004x + 5.05t)\\[/tex]
According to the above equation, k = 0.004 and ω = 5.05.
Using the following identities, we can find the period of the wave.
[tex]\omega = 2\pi f\\ f = 1/ T[/tex]
T = 1.25 s.
For the horizontal distance travelled by one period of time, x = λ.
[tex]\lambda = 2\pi / k = 2\pi / 0.004 = 1.57\times 10^3~m[/tex]
[tex]y(x = \lambda,t = T) = 0.033\cos(0.004*1.57*\10^3 + 5.05*1.25) = 0.033~m[/tex]
B) The wave number, k = 0.004 . The number of waves per second is the frequency, so f = 0.8.
C) The propagation speed of the wave is
[tex]v = \lambda f = 1.57\times 10^3 * 0.8 = 1.256\times 10^3~m/s[/tex]
The velocity of the wave is the derivative of the position function.
[tex]v(x,t) = \frac{dy(x,t)}{dt} = -(5.05\times 0.033)\sin(0.004x + 5.05t)[/tex]
The maximum velocity is when the derivative of the velocity function is equal to zero.
[tex]\frac{dv_y(x,t)}{dt} = -(5.05)^2(0.033)\cos(0.004*1.57\times 10^3 + 5.05t) = 0[/tex]
In order this to be zero, cosine term must be equal to zero.
[tex]0.004*1.57\times 10^3 + 5.05t = 5\pi /2\\t = 0.31~s[/tex]
The reason that cosine term is set to be 5π/2 is that time cannot be zero. For π/2 and 3π/2, t<0.
[tex]v(x=\lambda, t = 0.31) = -(5.05\times0.033)\sin(0.004\times 1.57\times 10^3 + 5.05\times 0.31) = -0.166~m/s[/tex]
