Question: (1 point) A spring is stretched 14 cm by a force of 9 N. (Note that by Hooke's law this means that F = kx

(1 point) A spring is stretched 14 cm by a force
(1 point) A spring is stretched 14 cm by a force of 9 N. (Note that by Hooke's law this means that F = kx where F = 9 N is force, c = 14 cm is displacement, and ( is the spring constant.) A mass of 8 kg is hung from the spring and also attached to a damper that exerts a force in the direction opposite to the direction of motion of the mass with magnitude proportional to the speed of the mass. The damper exerts a force of 9 N when the speed is 10 m/s. If the mass is pulled 20 cm below its equilibrium position and given an initial downward velocity of 20 cm/s, find the position u (in m) of the mass at any time t (in s). (Assume that position is measured upward from the equilibrium position.) u(t) = Find the quasifrequency / (in radians per second). Note: If you enter a decimal approximation, use at least seven digits after the decimal point. Find the ratio of / to the natural frequency of of the corresponding undamped system. w/ f = Note: If you enter a decimal approximation, use at least seven digits after the decimal point

Step by Step Solution

There are 3 Steps involved in it

1 Expert Approved Answer
Step: 1 Unlock blur-text-image
Question Has Been Solved by an Expert!

Get step-by-step solutions from verified subject matter experts

Step: 2 Unlock
Step: 3 Unlock

Students Have Also Explored These Related Mathematics Questions!