A sinusoidal wave of frequency 500 Hzhas a speed of 350 m/s . (a)How far apart are two points that differ in phase by$\mathbf{\pi }\mathbf{/}\mathbf{3}\mathbf{rad}$ ? (b)What is the phase difference between two displacements at a certain point at times 1.00 msapart?

• The frequency of the wave,$f=500Hz$
• The speed of the wave,$v=350m/s$
• The time period of the wave, $T=1.00ms$

The number of oscillations completed by the wave, per unit time interval is known as frequency. The distance between two consecutive crest or trough gives the wavelength of the wave. The product of wavelength of the wave and the frequency gives us the speed of the propagation of wave.

Formula:

The speed of the wave, $v=f\lambda$ (i)

The frequency of the wave, $f=\frac{1}{T}$ (ii)

Using equation (i), we can find the length of one cycle of the wave as given:

$$\begin{gathered} \lambda =\frac{v}{f} \\ =\frac{350m/s}{500s^{-1}} \\ =0.7m \\ =700mm \end{gathered}$$

A cycle is equivalent to $2\mathrm{\pi }\mathrm{rad}$so that $\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$ corresponds to one-sixth of a cycle.

The corresponding length, therefore is,

$$\begin{gathered} \frac{\lambda }{6}=\frac{700mm}{6} \\ =116.67mm \\ \approx 117mm \end{gathered}$$

Hence, the value of distance between the two points is 117 mm

From equation (ii), we can find the time for one cycle of oscillation as given:

$$\begin{gathered} T=\frac{1}{f} \\ =\frac{1}{500s^{-1}} \\ =2.00\times {10}^{-3}s \\ =2.00ms \end{gathered}$$

The interval 1.00 ms is half of T , thus corresponding to half of one cycle. Thus, the phase difference is given by

$\frac{1}{2}\times 2\mathrm{\pi }=\mathrm{\pi }\mathrm{rad}$

Hence, the value o the phase difference is$\mathrm{\pi }\mathrm{rad}$ .