An object in simple harmonic motion has amplitude 4.0{\ m cm} and frequency 3.6{

Question

An object in simple harmonic motion has amplitude 4.0{ m cm} and frequency 3.6{ m Hz} , and at t = 0;{ m s} it passes through the equilibrium point moving to the right. Write the function x(t) that describes the object's position. Write the function x(t) that describes the object's position. x(t) =( 4.0{ m cm} )cos((7.2pi ;{ m rad/s}) t -pi /2 ;{ m rad}) x(t) =(2.0{ m cm} )cos((7.2pi ; { m rad/s}) t -pi /2 ;{ m rad}) x(t) = (4.0{ m cm} )cos((3.6 ;{ m rad/s} )t) x(t) =(2.0{ m cm} )cos((7.2pi ; { m rad/s}) t)

Explanation / Answer

k = F/x = mg/x = 1.19(9.8)/0.103 = 113.223301 N/m w = sqrt(k/m) = sqrt(113/1.19) = 9.74 rad/s w = 2(pi) f = 2(pi)/T T =2(pi)/w= 2(pi)/(9.74) T = 0.66seconds please rate me

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