Question

On a wave with a wave velocity of $360m/s$ and a frequency of $500Hz$, the distance between two locations that are $60degrees$ out of phase is

• A. $0.72\mathrm{m}$
• B. $0.18\mathrm{m}$
• C. $0.12\mathrm{m}$
• D. $0.32\mathrm{m}$

Answer 1

The correct option is C

$0.12\mathrm{m}$

Step: 1 Given data:

Phase difference $\left(\mathrm{\theta }\right)={60}^{\mathrm{o}}$

Wave velocity $\left(\mathrm{v}\right)=360{\mathrm{ms}}^{-1}$

Frequency $\left(\mathrm{f}\right)=500\mathrm{Hz}$

Step: 2 Formula used:

The wavelength of the wave $\left(\mathrm{\lambda }\right)=\frac{\mathrm{v}}{\mathrm{f}}\dots \dots \dots \dots \dots \dots \left(1\right)$

$∆\varphi =\frac{2\mathrm{\pi }}{\lambda }\times ∆x\dots \dots \dots \dots \dots \left(2\right)$, where $\left(∆\varphi \right)=$phase difference, and $∆x=$path difference.

Step: 3 Calculation of wavelength:

Using the formula from equation $\left(1\right)$

$\mathrm{\lambda }=\frac{\mathrm{v}}{\mathrm{f}}=\frac{360}{500}=0.72\mathrm{m}$

Step 4: Calculating the phase difference:

Using the formula from equation $\left(2\right)$

Phase difference $\left(∆\varphi \right)=\frac{2\mathrm{\pi }}{\lambda }\times ∆x$

$\begin{array}{rcl}∆\varphi & =& \frac{\mathrm{\pi }}{3}rad\\ ∆x& =& \frac{\lambda ∆\varphi }{2\mathrm{\pi }}\\ & =& \frac{\lambda }{2\mathrm{\pi }}\times \frac{\mathrm{\pi }}{3}\\ & =& \frac{\mathrm{\pi }}{6}\\ & =& 0.12m\end{array}$

Hence option C is the correct answer.