Why Does Adding 2π to the Phase Angle Yield Equivalent Solutions in SHM?

[salman213]

In simple harmonic motion example there is a question

velocity max is given
velocity at time = 0 is given (which is not equal to maximum velocity --> not at equilibrium point at time=0)
graph is also given of the velocity function

IT asks to find the phase angle

in the solution they use the formula

v(t)=-vmaxsin(\omegat + \varphi)

using vmax and v(0) they solve for \varphi

then they add 2pie to it and say the answer is either x rad or 2pie + xrad (where x is the answer they found)

why do they add this 2pie (i know it has the same value but it can then be k2pie where k is any integer since x rad + 2pie or x rad + 4 pie or x rad + 6pie etc. will also give u the same answer)


any help?

 

[Shooting Star]

This is a HW question. Anyway, you've given some thought to it. You are right -- adding 2k*pi to any angle makes it the same.

Are you sure they are not saying (pi-x_rad)?

 

[salman213]

no that is also an answer yes but they say it is not the phase angle because if u use the angle u get from pie - x_rad that gives u the angle when the "slope" of the graph is increasing and therefore meaning the direction of the particle moving changes. IT is suppose to be going away from the equilibirum but if u say the phase angle is pie minus xrad then that means its is going towards the equilibrium point.

 

[Shooting Star]

Then it is the simple and correct answer that at phase omega+2n*pi, the particle is in the identical state. You cannot distinguish the state of the particle between any of these states, and that is why it's called to be in the same phase. (Actually, it had been my mistake not to explain to you sooner.)

Suppose a wave is moving. Then all the particles which are in the same phase along the wave are those that that have the angles as 2n*pi + omega, where n is any integer, +ve, -ve or zero. They have the same velocity in the same direction, and are at the same distance from the mean position