In some problems of simple harmonic motion, we take phase constant as zero, and in some, we do not....why?

Question

like we take it as (pie)/2 at extreme position, and sometimes, we write equations ignoring it totally.....how do i know when i gotta take it as zero....if its not specified in the problem?
i promise to select ur answer as the best if i am convinced.
thanks a lot for ur help....

Answer 1

The phase determines the position at t = 0. This obviously depends on what is described in the problem. These are, in general, called boundary conditions.

Bozo

Answer 2

The phase angle is included in order to take in account the starting point of simple harmonic motion.
It's just like displacement in normal motion

Answer 3

in my experience, if it is not specified. A zero phase is the norm. A modified phase is usually a solution to a problem that isnt working at zero.

Hope this helps

Answer 4

an hassle-free harmonic action is a vibratory action in a line between 2 factors. the value at the two the tip factors is 0, and maximum on the midsection. The action is pushed with tips from an elastic restoring stress proportional to the gap from the midsection. Acceleration being proportional to the stress, as a result it is often proportional to the gap from the midsection additionally and constantly is interior the adverse direction of the action, because of the fact the stress is a restoring stress, constantly attempting to entice the particle under action to the midsection or the equilibrium place. If a particle undergoes a action of uniform speed alongside a circle, its projection on a particular diameter undergoes an hassle-free harmonic action. think the radius vector of the particle on the circle makes an perspective theta = wt with a fastened radius and if the radius is of length A, ( the diameter being 2A) and if the fastened radius vector makes an perspective ? with fastened diameter, the radius vector of any component on the circle makes an perspective wt + ? with the fastened diameter and Acos(wt + ?) is the projection of the revolving radius vector on the fastened diameter. (wt + ?) is termed the 'area' of the shifting component on the circle and ? is termed the 'area perspective'. of course, (wt + ?) is the perspective defined with tips from the shifting component on the circle and A(wt + ?) is the arc coated with tips from it in time t, the gap of the projection on the diameter of the component shifting on the circle is displaced with tips from X = Acos(wt + ?) from the midsection. Now for looking speed of the projection component, differentiate it wrt X to get v =d X / d t= Awsin(wt + ?) = Awcos(wt + ?+ninety). it is defined as ' the version in area between displacement and speed of a particle executing SHM' is ninety tiers.A 2d time differentiation provides accelearation a = dv / dt = - Aw.wcos(wt + ?), that's having area distinction of one hundred eighty tiers with the area of displacement. Now a = dv / dt = - w.w.X, so, proportional to displacement and it is an hassle-free harmonic action with tips from definition!.