Question

Write the equation for a sinusoidal wave propagating in the negative \(x\) -direction with a speed of \(120 . \mathrm{m} / \mathrm{s}\), if a particle in the medium in which the wave is moving is observed to swing back and forth through a \(6.00-\mathrm{cm}\) range in \(4.00 \mathrm{~s}\). Assume that \(t=0\) is taken to be the instant when the particle is at \(y=0\) and that the particle moves in the positive \(y\) -direction immediately after \(t=0 .\)

Short answer

Answer: The equation representing the sinusoidal wave is: \(y(x, t) = 0.0300 \sin\left(\dfrac{\pi}{240} x + \dfrac{\pi}{2} t + \dfrac{\pi}{2}\right) m\).

Step-by-step solution

Determine the amplitude A

The particle oscillates through a range of 6 cm. As the range is from the minimum to maximum displacement of the particle, the amplitude \(A\) can be calculated. Since the amplitude is half the range, \(A = \dfrac{6.00 cm}{2} = 3.00 cm = 0.0300 m\).

Determine the angular frequency ω

The particle takes 4 seconds to complete one oscillation. Now calculate the angular frequency (\(\omega\)) from the period \(T\): \(T = 4.00 s\) \(\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{4.00 s} = \dfrac{\pi}{2} \mathrm{rad/s}\).

Determine the phase shift φ

As the particle moves in the positive \(y\)-direction immediately after t=0, it means that the wave begins a quarter of a cycle after the maximum positive displacement. Hence, the phase shift (\(\phi\)) is given by: \(\phi = \dfrac{1}{4} (2\pi) = \dfrac{\pi}{2} \mathrm{rad}\).

Determine the angular wave number k

The speed of propagation \(v\) can be related to the angular wave number (\(k\)) and the angular frequency (\(\omega\)): \(v = \dfrac{\omega}{k}\). Now solve for k: \(k = \dfrac{\omega}{v} = \dfrac{\dfrac{\pi}{2} \mathrm{rad/s}}{120 \mathrm{m/s}} = \dfrac{\pi}{240} \mathrm{rad/m}\).

Write the wave equation

Putting all the values in the wave equation: \(y(x, t) = A\sin(kx + \omega t + \phi)\) \(y(x, t) = 0.0300 \sin\left(\dfrac{\pi}{240} x + \dfrac{\pi}{2} t + \dfrac{\pi}{2}\right) m\).