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A mass of 1kg stretches a spring by 32cm. The damping constant is c=0. Exterbal vibrations...

A mass of 1kg stretches a spring by 32cm. The damping constant is c=0. Exterbal vibrations create a force of F(t)= 4 sin 3t Netwons, setting the spring in motion from its equilibrium position with zero velocity. What is the coefficient of sin 3t of the steady-state solution? Use g=9.8 m/s^2. Express your answe is two decimal places.
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Answer #1

Spring Constant is

K=mg/x =1*9.8/0.32 =30.625 N/m

The differential equation of the system is of the form is

Let assume particular integral is

substitute in differential equation is

compare sin terms

-9A+30.625A=4

A=0.18497 =0.18

Compare cos terms

-9B+30.625B=0

B=0

Therefore

Therefore answer is 0.18

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