Question

A sinusoidal wave is described by y(x,t) = 3.0 sin (5.95t-4.20x)cm
where x is the position along the wave propagation. Determine the amplitude, wave
number, wavelength, frequency and velocity of the wave. (express in answer in meters not centimeter)

Answer 1

Answer on Question 73645, Physics / Mechanics | Relativity

Question

A sinusoidal wave is described by $y(x,t) = 3.0 \sin (5.95t - 4.20x)cm$ where $x$ is the position along the wave propagation. Determine the amplitude, wave number, wavelength, frequency and velocity of the wave. (express in answer in meters not centimeter)

Solution. In the general case a sinusoidal wave is described by

$$y(x,t) = y_0 \sin(\omega t - kx)$$

where $y_0$ is the amplitude, $\omega$ is the angular frequency, is the wave number

In this problem we have

$$y(x,t) = 3 \sin(5.95t - 4.20x)$$

Then we get:

the amplitude of the wave is $y_0 = 3.0cm = 0.03 \, m$

the wave number of the wave is $k = 4.20 \, cm^{-1} = 420 \, m^{-1}$

the wavelength of the wave is $\lambda = \frac{2\pi}{k} = \frac{2\pi}{420 \, m^{-1}} = 0.015 \, m$

the angular frequency of the wave is $\omega = 5.95 \, s^{-1}$

the frequency of the wave is $f = \frac{\omega}{2\pi} = \frac{5.95 \, s^{-1}}{2\pi} = 0.95 \, \text{Hz}$

the velocity of the wave is $v = \frac{\omega}{k} = \frac{5.95 \, s^{-1}}{420 \, m^{-1}} = 0.014 \, m/s$

Answer: the amplitude, wave number, wavelength, frequency and velocity of the wave are

$$y_0 = 0.03 \, m$$ $$k = 420 \, m^{-1}$$ $$\lambda = 0.015 \, m$$ $$f = 0.95 \, \text{Hz}$$ $$v = 0.014 \, m/s$$

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