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A sinusoidal wave traveling in the +𝑥 direction (to the right) has an amplitude of 15.0 cm, a wavelength of 10.0 cm and a frequency of 20.0 Hz. At t = 0, a particle at x = 0 has a displacement of 15.0 cm.(a) Write an expression for the wave function, y(x, t).20:39b) Determine the speed and acceleration at t = 0.500 s for the particle on the wave located at x = 5.0 cm.20:40

Question

A sinusoidal wave traveling in the +𝑥 direction (to the right) has an amplitude of 15.0 cm, a wavelength of 10.0 cm and a frequency of 20.0 Hz. At t = 0, a particle at x = 0 has a displacement of 15.0 cm.(a) Write an expression for the wave function, y(x, t).20:39b) Determine the speed and acceleration at t = 0.500 s for the particle on the wave located at x = 5.0 cm.20:40

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Solution

(a) The general equation for a wave moving in the positive x direction is given by:

y(x, t) = A sin(kx - wt + φ)

where: A = amplitude = 15.0 cm k = wave number = 2π/λ = 2π/10.0 cm = 0.2π cm^-1 w = angular frequency = 2πf = 2π*20.0 Hz = 40π rad/s φ = phase constant

Given that at t = 0, a particle at x = 0 has a displacement of 15.0 cm, the phase constant φ = 0 (since sin(0) = 0 and the amplitude is 15 cm). So, the wave function is:

y(x, t) = 15.0 cm sin(0.2πx - 40πt)

(b) The speed v of the wave is given by the equation v = λf = 10.0 cm * 20.0 Hz = 200 cm/s.

The acceleration a of a particle at a point on the wave is given by the second derivative of the wave function with respect to time:

a = -A*w^2 sin(kx - wt + φ)

At t = 0.500 s and x = 5.0 cm, this becomes:

a = -15.0 cm * (40π rad/s)^2 * sin(0.2π5.0 cm - 40π0.500 s) = -15.0 cm * (40π rad/s)^2 * sin(π - 20π rad) = -15.0 cm * (40π rad/s)^2 * sin(-19π rad) = 15.0 cm * (40π rad/s)^2 * sin(19π rad) = 15.0 cm * (40π rad/s)^2 * 0 = 0 cm/s^2

So, the speed of the wave is 200 cm/s and the acceleration of the particle at t = 0.500 s and x = 5.0 cm is 0 cm/s^2.

This problem has been solved

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