Answer to Question #1836 in Mechanics | Relativity for nirmala

Question

A wave is described by the following equation:
y = (1.0 nm) sin 2Ï€[t/.01s-x/2cm]

(i) Calculate the time period and the wavelength of the wave.

(ii) Obtain the expression for the velocity of the particles.

(iii) Calculate the speed of the particles at x = 3.0 cm, 5.0 cm and 7.0 cm at t = 0.01 s.

(iv) Calculate speed of the particles at x = 1.0 cm, at t = 0.011 s, 0.012 s and 0.013s.

Answer 1

The time period& of the wave:& T = 2π / (2π/0.01) = 100 s,& the wavelength is& λ = 2π / (2π/2) = 2 cm.
The velocity is dy/dt = 200π cos 2π[t/.01-x/2] [nm/s]
& V(x=3.0cm, t= 0.01 s) = 200π cos 2π(.01/.01-3/2) = -627.999 nm/s
& V(x=5.0cm, t= 0.01 s) =& -627.992 nm/s
& V(x=7.0cm, t= 0.01 s) =& -627.980 nm/s

& V(x=1.0cm, t= 0.011 s) = 200π cos 2π(.011/.01-1/2) = -508.767 nm/s
& V(x=1.0cm, t= 0.012 s) = 200π cos 2π(.012/.01-1/2) = -195.394 nm/s
& V(x=1.0cm, t= 0.013 s) = 200π cos 2π(.013/.01-1/2) = 192.540 nm/s