A wave of frequency M sec-1 has a velocity of 4M m/sec.
(a) How far apart are two points whose displacements are (饾懘饾煇)饾憸. apart in phase?
(b) At a given point, what is the phase difference between two displacements occurring at times separated by M/1000 sec
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To solve the given questions, we need to understand the concepts of wavelength and phase difference.
(a) The distance between two points with displacements (饾懘^2)饾憸. apart in phase is equal to one wavelength (位) of the wave.
We know that the velocity (v) of a wave is given by the product of its frequency (f) and wavelength (位). So, v = f * 位.
From the given information, the frequency (f) of the wave is M sec^-1 and the velocity (v) is 4M m/sec.
So, we can write the equation as: 4M = M * 位
Simplifying, we get: 位 = 4
Therefore, the two points whose displacements are (饾懘^2)饾憸. apart in phase are 4 meters apart.
(b) The phase difference (螖饾湋) between two displacements occurring at times separated by M/1000 sec can be calculated using the formula:
螖饾湋 = 2蟺 * (螖t / T)
Where 螖t is the time interval between the two displacements and T is the period of the wave.
The period is the reciprocal of the frequency, so T = 1/f.
Given that the time interval is M/1000 sec and the frequency is M sec^-1, we have:
T = 1/M and 螖t = M/1000
Substituting these values into the formula, we get:
螖饾湋 = 2蟺 * (M/1000) * (1/M)
Simplifying, we get:
螖饾湋 = 2蟺/1000
Therefore, the phase difference between the two displacements is 2蟺/1000 radians.