Question

Consider the waveform expression. y(x, t) = Ymsin (3.38 + 261t + 0.129x) The transverse displacement (y) of a wave is given as a function of position (x in meters) and time (t in seconds) by the expression. Determine the wavelength, frequency, period, and phase constant of this waveform. a= meters f = Hertz T = seconds ΦΟ = radians

Answer 1

y = ym sin(3.38 + 261t + 0.129 x)

y = A sin( phi0 + w t + k x )

(A) wavelength = 2 pi / k = 2 pi / 0.129 = 48.7 m

(B) f = w / 2pi = 261 / 2pi = 41.5 Hz

(C) T = 1/f = 0.024 s

(D) phi0 = 3.38 rad

Answer 2

A general traveling wave can be expressed as

$$y=y_{\text{m}}\sin (kx\pm \omega t+{\varphi }_0)$$

where ?$k$ is the wavenumber, ?$\omega$ is the angular frequency, and ?0${\varphi }_0$ is called the phase constant. Alternatively, a cosine function could be used in the place of the sine function (which would alter the value of ?0${\varphi }_0$, and nothing else). The order of the three terms is unimportant, but ?$k$ is always the coefficient of the position (?$x$) and ?$\omega$ is always the coefficient of the time (?$t$). Therefore, the wavenumber, angular frequency, and phase constant are

$$k=0.197\text{ radians/m}\omega =909\text{ radians/s }\varphi =5.36\text{ radians}$$

The wavelength is related to ?$k$, whereas the temporal properties (frequency and period) are related to ?$\omega$. To see how, start with the fact that sine (or cosine) repeats every 2?$2\pi$ radians. The phase constant does not change, so at a fixed time, the distance over which the waveform repeats is called the wavelength. So, putting the wavelength (?$\lambda$) in place of the distance (?$x$), the product of the wavenumber and the wavelength must be equal to 2?$2\pi$ radians.

$$\begin{array}{rl}k\lambda & =2\pi \text{ radians}\\ \lambda & =\frac{2\pi \text{ radians}}{k}\\ & =\frac{2\pi \text{ radians}}{0.197\text{ radians}/\text{m}}\\ & =31.9\text{ m}\end{array}$$

Conversely, at a fixed position, the time over which the waveform repeats is called the period. Putting the period (?$T$) in place of the time (?$t$), the product of the angular frequency and the period must be equal to 2?$2\pi$ radians.

$$\begin{array}{rl}\omega T& =2\pi \text{ radians}\\ T& =\frac{2\pi \text{ radians}}{\omega }\\ & =\frac{2\pi \text{ radians}}{909\text{ radians}/\text{s}}\\ & =0.00691\text{ s}\end{array}$$

The frequency (?$f$) is just the inverse of the period, and it expresses how many oscillations a fixed position goes through per unit time.

?=1?=?2? radians=909 radians/s2? radians=145 s−1=145 Hz