Question

A traveling wave is described by the equation: TT y(x, t) = (0.35 cm) sin ( 3173 107t + 4 Find: a. the wavelength of the wave; b. the period of the wave; c. the speed and direction of travel of the wave; Now consider the particle at x = 10 cm. d. Find the vertical displacement of the particle t = 0. e. Find the maximum transverse speed and acceleration of the particle. If the oscillator that is generating the wave is at x = 0, f. what is its displacement and the direction of the oscillator's motion at t = 0?

Answer 1

A travelling wave is described by the equation,

Y(x, t) = (0.35 cm ) sin(3πx - 10πt + π/4)

The general equation for a travelling wave is

Y(x, t) = (A cm ) sin(kx - ωt + ɸ)

We get by comparing the two equations,

Amplitude of the wave, A = 0.35 cm

Angular frequency of the wave, ω = 10π rad/s

Angular wave number of the wave, k = 3π rad/cm

a.

wavelength of the wave, λ = 2π/k = 2π/3π = 0.67 cm = 0.0067 m

b.

period of the wave, T = 2π/ ω = 2π/10π = 0.2 s

c.

speed of the wave, v = λ/T = 0.67/0.2 = 3.4 cm/s = 0.034 m/s

for a particle at x = 10 cm

d.

at time t = o

Y(x, t) = (0.35 cm ) sin( - 10πx10+ π/4) = 0.25 cm = 0.0025 m

e.

the maximum speed of the particle, V_max = Aω = 0.35 x 10π = 11.0 cm/s = 0.11 m/s

the maximum acceleration of the particle, a_max = Aω² = 0.35 x 10π x 10π = 345 cm/s = 3.45 m/s

f.

at x = 0 and t = 0

Y(x, t) = (0.35 cm ) sin(π/4) = 0.25 cm = 0.0025 m