Question

A wave of frequency $500Hz$ has a velocity of ${360}^{°}m{s}^{-1}.$ Calculate the distance between two points that are ${60}^{°}$ out of phase

• A. $12cm$
• B. $18cm$
• C. $17cm$
• D. $25cm$

Answer 1

The correct option is A

$12cm$

Step 1. Given Data,

Assume, $2\mathrm{\pi }={360}^{°}$

Frequency $f=500Hz$

Phase difference $\varphi ={60}^{°}$

Velocity $v=360m{s}^{-1}$

We have to calculate difference between two points $∆x.$

Step 2. Calculate the wavelength of a wave of frequency $500Hz$,

Wavelength = Velocity $/$ frequency

$$\begin{gathered} \lambda =\frac{v}{f} \\ \lambda =\frac{360m{s}^{-1}}{500Hz} \\ \lambda =0·72m \\ \lambda =72cm \end{gathered}$$

Step 3. Calculate the distance between two points that are ${60}^{°}$ out of phase,

We know that,

Phase difference $=2\mathrm{\pi }\times$Distance between two points $/$wavelength

$\Rightarrow$ Distance between two points $=($Phase difference $\times$ wavelength$)/2\mathrm{\pi }$

$$\begin{gathered} ∆x=\frac{{60}^{°}\times 72cm}{{360}^{°}} \\ ∆x=12cm \end{gathered}$$

Hence option A is correct.