Question

A wave of frequency $500\,Hz$ has a wave velocity of $350\,m\,{s^{ - 1}}$ .

1. Find the distance between the two points which has ${60^0}$ out of phase.
2. Find the phase difference between two displacements at a certain point of time ${10^{ - 3}}\,s$ apart?

Answer 1

In this question, we will first calculate the wavelength of the wave from the velocity and the frequency already given to us. Then we shall use the relation that a phase difference of $2\pi$ is equivalent to a path difference of $\lambda$ and calculate the path difference by substituting the required values in the formula $x = \dfrac{{\phi \lambda }}{{2\pi }}$ . Further we shall calculate the distance travelled by the wave using the time given to us. Again, making proper substitutions, we will get our answer.

Complete step by step answer:
The velocity of the wave is given to be $V = 350\,m\,{s^{ - 1}}$
Also, frequency of the wave is given to be $\upsilon = 500\,Hz$
The wavelength, frequency and the velocity of the wave are related as $V = \lambda \upsilon$ where V is the velocity of the wave, $\lambda$ is the wavelength of the wave and $\upsilon$ is the frequency of the wave.
Substituting the known values, we get,
$\dfrac{{350}}{{500}} = \lambda$
$\Rightarrow \lambda = 0.7\,m$

1. We are supposed to calculate the distance between the two points given that they have a phase of ${60^0}$
   The same phase difference expressed in radians is equal to $\phi = \dfrac{\pi }{3}$
   A phase difference of $2\pi$ is equivalent to a path difference of $\lambda$
   Hence, we can say that if the path difference is $x$ for a phase difference of $\phi = \dfrac{\pi }{3}$
   $\phi = \dfrac{{2\pi x}}{\lambda }$
   This can be rewritten as
   $x = \dfrac{{\phi \lambda }}{{2\pi }}$
   Substituting the known values, we get,
   $x = \dfrac{{\pi \times 0.7}}{{6\pi }}$
   $\Rightarrow x = 0.116\,m$
2. The velocity of the wave is given to be $V = 350\,m\,{s^{ - 1}}$
   The time taken is given as $t = {10^{ - 3\,}}\,s$
   So, the distance travelled by the wave is equal to $x = vt$ where x is the distance travelled.
   Substituting in the equation, we get,
   $x = 350 \times {10^{ - 3}}$
   $\Rightarrow x = 0.35\,m$
   So, now we know the path difference and we are supposed to calculate the phase difference.
   Using the same relation as used before, we have,
   $\phi = \dfrac{{2\pi x}}{\lambda }$
   Substituting the values, we get,
   $\phi = \dfrac{{2\pi \times 0.35}}{{0.7}}$
   $\Rightarrow \phi = \pi$

Note: The phase difference refers to the magnitude of the difference in the phase. It can be both leading or lagging. It should be specified which quantity is leading and which is lagging. In this particular question, we didn’t specify the lead or lag because we were supposed to deal with quantities that could be calculated without knowing whether it is a lead or lag.