Two simple harmonic motions are represented by y_(1) = 10 sin"" (pi)/(4) (12t+1)" and " y_(2) = 5(sin"" 3pi t+ sqrt(3) cos 3pi t). Find out the ratio of their amplitudes. What are the time period of two motions.

Two simple harmonic motions are represented by y_(1) = 10 sin"" (pi)/(

1.	Two simple harmonic motions are represented by y_(1) = 10 sin"" (pi)/(4) (12t+1)" and " y_(2) = 5(sin"" 3pi t+ sqrt(3) cos 3pi t). Find out the ratio of their amplitudes. What are the time period of two motions.
Answer» Solution :`y_(1) = 10sin (pi)/(4) (12t+ 1)` COMPARING `y_(1) = 10sin (3pi t+ (pi)/(4))` with `y= A_(1) SIN (omega_(1) t+phi)` Amplitude `A_(1) = 10 cm " and "omega =3pi RADS^(-1)` `T = (2PI)/(omega) = (2pi)/(3pi) = (2)/(3)s` and `y_(2) = 5 (sin 3pi t + sqrt(3) cos 3 pi t)` `= 5xx2 [(1)/(2) sin3pi t + sqrt((3)/(2)) (cos 3 pi t)]` `y_(2) = 2XX 5[cos phi sin 3 pi t +sin phi (cos 3pi t)]` Comparing `Y_(2) = 10 [sin (3pi + phi)]` with `y= A_(2) sin (omega_(2) t +phi)` Amplitude `A_(2) = 10 cm" and "omega_(2) 3pi rads^(-1)` `therefore T_(2) = (2pi)/(omega_2) = (2pi)/(3pi) = (2)/(3)" second "` Ratio of amplitudes `(A_1)/(A_2) = (10)/(10)= 1` and `T_(1) = T_(2) = (2)/(3)" second "`.