Two particle executing S.H.M. of same amplitude of 20 cm with same period along the same line about same equilibrium position. The maximum distance between the two is 20 cm. Their phase difference in radian is equal to?

–1 vote

Two particle executing S.H.M. of same amplitude of 20 cm with same period along the same line about same equilibrium position. The maximum distance between the two is 20 cm. Their phase difference in radian is equal to?

The phase difference can be $\pi /2$ or $3\pi /2$

@Navin, this is an answer, not a comment. Please post as an answer and provide an explanation. Then other users can vote for it, and Anandit can accept it.

-1 Welcome to the site, Anandit . Please note that we expect you to show some effort to solve the problem yourself, and to explain what difficulty you are having.See http://physicsproblems.nfshost.com//?qa=116/scope-%26-guidelines .

Your problem might be solved using the answer to http://physicsproblems.nfshost.com//?qa=1856/phase-difference-between-the-particles&show=1856#q1856

0 votes

This is essentially the same question as Phase difference between the particles.

METHOD 1 :

The phase difference between the particles P and Q remains the same, but the separation changes. At maximum separation they are moving in the same direction with the same speed, and they are positioned symmetrically about equilibrium position O.

If P and Q are moving in opposite directions then they will get further apart or closer together, so they cannot be at their maximum separation.

Suppose that, at some instant, P and Q are moving in the same direction. When the separation is maximum they are moving with the same speed. If the speed of one was greater than the other, they would get closer together of further apart.

For both P and Q, speed decreases with distance from O in the same manner. Because they have the same amplitude, whichever is closer to O moves faster. They will have the same speed when they are the same distance from O, on opposite sides of O. So at maximum separation they will be positioned symmetrically about O.

We are told that the maximum separation is equal to the amplitude $A$. So when the separation is $A$ then each particle is $\frac12 A$ from O. The phase of each is then $\phi$ where

$A\sin\phi=\pm \frac12 A$

$\sin\phi=\pm \frac12=\sin(30, 150, 210, 330^{\circ})$

$\phi=30, 150, 210, 330^{\circ}$.

So the phase difference at maximum separation (and at all separations) is $2\phi=60, 300^{\circ} = \pi/3, 5\pi/3$ rad. (The phase difference cannot be more than $360^{\circ}$.)

METHOD 2 :

Suppose the displacements of the two particles are given by

$x_1=A\sin(\omega t+\phi)$

$x_2=A\sin(\omega t)$

where $A=20cm$ is the amplitude, $\omega$ is angular frequency (which is also the same), and $\phi$ is the unknown phase difference.

The distance between the particles is given by

$|x_1-x_2|=|2A\cos(\frac{\omega t+\phi+\omega t}{2})\sin(\frac{\omega t+\phi - \omega t}{2})|=|2A\cos(\omega t+\frac12 \phi)\sin(\frac12 \phi)|$.

Now $\sin(\frac12 \phi)$ is a constant, so the maximum possible value of $|x_1-x_2|$ occurs when $\cos(\omega t+\frac12 \phi)=\pm 1$. The distance between the particles equals $A=20cm$ when

$ \pm 2A\sin(\frac12 \phi) = A$

$\sin(\frac12 \phi)=\pm \frac12 = \sin(30, 150, 210, 330^{\circ})$

$\phi=60, 300, 420, 660^{\circ}$.

The phase difference is $60^{\circ}=\pi/3$ rad or $300^{\circ}=5\pi/3$ rad.

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