
Two particles execute S.H.M of same amplitude and frequency along the same straight line from the same mean position. They cross one another without collision, when going in the opposite direction, each time their displacement is half of their amplitude. The phase-difference between them is:
A) ${0^o}$
B) ${120^o}$
C) ${180^o}$
D) ${135^o}$
Answer
134.4k+ views
Hint: Here it is given that two particles execute S.H.M and their amplitude and frequency are also the same. So, first assume such particles executing S.H.M in the form of sine and cosine waves. After assuming the motion of the particles focus on the point that each time their displacement is half of their amplitudes. This will help you determine the initial phase and then as they move in the opposite direction after collision so accordingly the phase difference would also be calculated.
Complete step by step solution:
Here it is given in the question that these two particles execute S.H.M and both the particles have same amplitude and frequency, so the equation would be given by,
${x_1} = {x_m}\cos \omega t$
${x_2} = {x_m}\cos (\omega t + \phi )$
It is given that these two particles cross paths with each other. Also, their displacement is half of their amplitudes, so here ${x_1} = \dfrac{{{x_m}}}{2}$
As, ${x_1} = {x_m}\cos \omega t$
So, equating the both expressions of ${x_1}$ we have,
$\dfrac{{{x_m}}}{2} = {x_m}\cos \omega t$
After cancelling ${x_m}$ from both the sides we have,
$\cos \omega t = \dfrac{1}{2}$
So, we have $\omega t = \dfrac{\pi }{3}$
Now similarly for ${x_2}$ we are given that they are moving in different directions after collision, so the closest value of $\omega t + \phi = \dfrac{{2\pi }}{3}$
Putting the value of $\omega t = \dfrac{\pi }{3}$
We have,
$\dfrac{\pi }{3} + \phi = \dfrac{{2\pi }}{3}$
On solving for the value of $\phi $ we have,
$\phi = \dfrac{\pi }{3}$
In degrees this angle would be equal to ${120^o}$ .
So here the correct answer is option B that is ${120^o}$.
Note: It is important to note the concept of Phase Difference. It is used to describe the difference in degrees or difference in radians when two or more alternating quantities reach their maximum or zero value. In other words phase difference is the lateral difference between two or more waveforms along a common axis and sinusoidal waveforms of the same frequency can have a phase difference.
Complete step by step solution:
Here it is given in the question that these two particles execute S.H.M and both the particles have same amplitude and frequency, so the equation would be given by,
${x_1} = {x_m}\cos \omega t$
${x_2} = {x_m}\cos (\omega t + \phi )$
It is given that these two particles cross paths with each other. Also, their displacement is half of their amplitudes, so here ${x_1} = \dfrac{{{x_m}}}{2}$
As, ${x_1} = {x_m}\cos \omega t$
So, equating the both expressions of ${x_1}$ we have,
$\dfrac{{{x_m}}}{2} = {x_m}\cos \omega t$
After cancelling ${x_m}$ from both the sides we have,
$\cos \omega t = \dfrac{1}{2}$
So, we have $\omega t = \dfrac{\pi }{3}$
Now similarly for ${x_2}$ we are given that they are moving in different directions after collision, so the closest value of $\omega t + \phi = \dfrac{{2\pi }}{3}$
Putting the value of $\omega t = \dfrac{\pi }{3}$
We have,
$\dfrac{\pi }{3} + \phi = \dfrac{{2\pi }}{3}$
On solving for the value of $\phi $ we have,
$\phi = \dfrac{\pi }{3}$
In degrees this angle would be equal to ${120^o}$ .
So here the correct answer is option B that is ${120^o}$.
Note: It is important to note the concept of Phase Difference. It is used to describe the difference in degrees or difference in radians when two or more alternating quantities reach their maximum or zero value. In other words phase difference is the lateral difference between two or more waveforms along a common axis and sinusoidal waveforms of the same frequency can have a phase difference.
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