Question

A sinusoidal wave is described by
y (x, t) = 4.0 sin (4.20x − 5.95t) cm
where x is the position along the wave propagation. Determine the amplitude, wave
number, wavelength, frequency and velocity of the wave.

Answer 1

Task. A sinusoidal wave is described by $y(x,t)=4.0\sin(4.20x-5.95t)$ cm, where $x$ is the position along the wave propagation. Determine the amplitude, wave number, wavelength, frequency and velocity of the wave.

Solution. In general, a sinusoidal wave has the following equation:

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

where $A$ is the amplitude, $k$ is the wave number and $\omega$ is the angular frequency.

There is no information about the measures of $x$ and $t$. Therefore let us denote the measure of $x$ by $L$ and the measure of time $t$ by $T$. Then

$A=4.0\ cm,\qquad k=4.20\ L^{-1},\qquad\omega=5.95\ T^{-1}.$

Recall that

$k=\frac{2\pi}{\lambda},\qquad\omega=2\pi f,$

where $\lambda$ is the wavelength, and $f$ is the frequency. Therefore

$\lambda=\frac{2\pi}{k}=\frac{2\cdot 3.14}{4.20}\approx 1.50\ L,$

$f=\frac{\omega}{2\pi}=\frac{5.95}{2\cdot 3.14}\approx 0.95\ T^{-1}.$

The velocity of the wave is given by the formula:

$v=\frac{\omega}{k}=\frac{5.95}{4.20}\approx 1.42\ L/T.$

Answer.

amplitude: $A=4.0\ cm$,

wave number: $k=4.20\ L^{-1}$,

wavelength: $\lambda=1.50\ L$,

frequency: $f=0.95\ T^{-1}$,

velocity: $v=1.42\ L/T$.