Is the phase constant of a pendulum undergoing simple harmonic motion determined by the angle at t=0?

Question

Is the phase constant of a pendulum undergoing simple harmonic motion determined by the angle at t=0?

Answer 1

To assign a phase, you need to establish a benchmark. It is often useful and convenient to set angle =0 to the vertical line associated with the pendulum at its lowest point, but you could use any line as your benchmark. Similarly, you can use the angle of the pendulum at any time; but once you do this, you need to be consistent. It is very convenient to use the angle of the pendulum with respect to some chosen benchmark at t=0, but it is not necessary.

Answer 2

A phase measurement implies a comparison between two angles. The pendulum is one, what is the other?

Answer 3

you're searching for phi and A such that FOR ALL VALUES of t: Acos(omega t + phi) = Ccos(omega t) + Ssin(omega t). So %. any 2 distinctive values of t. a solid start up is t=0: A cos( omega.0 + phi) = Ccos (omega.0) + S sin(omega 0) => A cos( phi ) = C cos (0 ) + X sin (0) => A cos( phi ) = C......... Equation a million in addition set t=pi/(2 omega) (as a manner to make omega.t = pi/2): A cos(pi/2 + phi) = C cos(pi/2) + S sin(pi/2) =>-A sin(phi) = S ........Equation 2 {because of the fact cos pi/2 = 0 and sin pi/2 = a million and cos(something+pi/2) = -sin(something) } you may now sq. equations a million and a pair of and upload to get A: A squared cos squared phi + Asquared sin squared phi = C squared + S squared So A squared = C squared + S squared { i.e. Pythagoras} So A = sqrt( C**2 + S**2 ) Similary you may divide equation 2 by making use of equation a million and arctan to get phi: -sin(phi)/cos(phi) = S/C => tan(phi) = -S/C => phi = arctan (-S/C) { or “tan to the minus a million S over C”} in case you don’t like algebra, draw an exact angled triangle with facets C and S and hypotenuse A and an perspective of phi between the the C and S facets. (yet pay attention of the sign of phi.) Then seem up an digital engineering e book on phasors and sophisticated numbers (which makes the mathematics greater trouble-free). in case you somewhat like algebra and trigonometry and don’t like placing particular values of t you may play around with hideous expressions like A cos( omega t + tan to the minus one in all (-S/C) ) to examine the respond (don’t forget cos(x+y) = cos x cos y – sin x sin y). yet you nevertheless land up drawing little triangles contained in the margin to artwork out that cos(tan to the minus a million of (S/C)) = C/sqrt(C**2+S**2). So all of it gets right down to Pythagoras.