A mass of 514 g stretches a spring by 9.9 cm. The damping constant is c...

Transcribed Image Text:

A mass of 514 g stretches a spring by 9.9 cm. The damping constant is c = 0.42. External vibrations create a force of F(t)= 0.3 cos 3t Newtons, setting the spring in motion from its equilibrium position with zero velocity. What is the imaginary part v, of the complex root of the homogeneous equation? Use g = 9.8- -. Express your answer in two decimal places. A mass of 514 g stretches a spring by 9.9 cm. The damping constant is c = 0.42. External vibrations create a force of F(t)= 0.3 cos 3t Newtons, setting the spring in motion from its equilibrium position with zero velocity. What is the imaginary part v, of the complex root of the homogeneous equation? Use g = 9.8- -. Express your answer in two decimal places.

Answer rating: 100% (QA)

Solution Sol Given mass Cm 514 9 0514 kg stretches spring 99 am 00 99 m damping ... View the full answer