A SHM motion is represented by `y=3sin2 pi t+4sin(2 pi t+(pi)/(6))+4sin(2 pi t+(5 pi)/(6))cm.` The amplitude and time period of motion are ?

period of sin(2 pi t+(pi)/(6))

A simple harmonic motion is represented by : y = 5(sin 3pi t + sqrt(3) cos 3pi t)cm The amplitude and time period of the motion are :

sin pi/3 = 2sin pi/6 cos pi/6

A simple harmonic motion is represented by y = 8 sin 10 pi t + 6 cos 10 pit , where is in cm and t is in sec. Calculate the amplitude time period and intial phase.

Two simple harmonic motions are represented by y_(1)=5 [sin 2 pi t + sqrt(3)cos 2 pi t] and y_(2) = 5 sin (2pit+(pi)/(4)) The ratio of their amplitudes is

A simple wave motion represented by y=5(sin 4pi t+sqrt3 cos 4pit) . Its amplitude is

A simple harmonic wave having an amplitude a and time period T is represented by the equation y = 5"sin"pi(t + 4)m . Then the value of amplitude (a) in (m) and time period (T) in second are

A linear SHM is represented by x=5sqrt(2) ("sin" 2pi t+ "cos" 2pi t) cm. What is the amplitude of SHM ?

Two simple harmonic motions are represented by the equations : x_1 = 5 sin (2 pi t + pi//4 ) , x_2 = 5^(2) (sin2 pi t + cos2 pi t) What is the ratio of their amplitudes ?

sin ((pi) / (5)) sin ((2 pi) / (5)) sin ((3 pi) / (5)) sin ((4 pi) / (5)) =