Two parallel S.H.M. s are given by , `x _1 = 20 sin (8 (pi) t)` and `x_2 = 10 sin [8 (pi) t + (pi)/ 6]` Find the resultant amplitude and initial phase of resultant S.H.M.

Two parallel S.H.M's are given by x_1 = 20sin8pit and x_2 =10sin[8pi t +frac(pi)(6)] . Find the resultant amplitude and phase.

Two parallel S.H.M.s have equations y_1=A_1sin(omegat+2pi) and y_2=A_2sin(omegat+4pi) . The amplitude of the resultant motion is

Two waves are represented by y_(1)= a sin (omega t + ( pi)/(6)) and y_(2) = a cos omega t . What will be their resultant amplitude

Two parallel S.H.M's represented by x_1 = 5sin [4pi t+frac(pi)(3)] cm x_2 = 3sin [4pi t _+frac (pi)(4)]cm are superposed on a particle.Determine the amplitude and epoch of the resultant S.H.M.

Two particles are executing S.H.M. according to the equations x_1 = 6 sin (10 pi t+ pi/3) and x_2 = 6 sin (8 pi t + pi/4) .Then the phase difference between the first and second particle at t = 0.5 s will be

Two simple harmonic motions are given by y_(1) = a sin [((pi)/(2))t + phi] and y_(2) = b sin [((2pi)/( 3))t + phi] . The phase difference between these after 1 s is

The displacement of a particle linear S.H.M. is given by x = 6 sin [3(pi)t + 5(pi) /6] meter. Find amplitude, frequency and the phase constant of the motion .

The S.H.M.s of two particles are given by y_(1)=10 "sin" [2pi t +(pi)/(6)] " and " y_(2) =5["sin" 2pi t + sqrt(3) "cos" 2 pi t] The ratio of their amplitudes is