A 4-kg mass is attached to a spring hanging from the ceiling, thereby causing the spring to stretch 2.45 m upon coming to rest at equilibrium. At time t=0, an external force of F(t) = 4 cos t N is applied to the system. The damping constant for the system is 5 N-sec/m. Determine the steady-state solution for the system. The steady-state solution is y(t) =.

A 4-kg mass is attached to a spring hanging from the ceiling, thereby causing the spring to stretch 2.45 m upon coming to rest at equilibrium. At time t=0, an external force of F(t) = 4 cos t N is applied to the system. The damping constant for the system is 5 N-sec/m. Determine the steady-state solution for the system. The steady-state solution is y(t) =.

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

See similar textbooks

Similar questions

A 20 kg weight is attached to a spring with constant k = 320 kg/m and subjected to an external force F(t) = 1200 cos(2t). The weight begins at rest in its equilibrium position. Find its displacement for t> 0, with y(t) measured positive upwards. y(t) =
灣彩 A 9.1kg body is suspended from a spring of constant k = 2.0kN/m. At time t = 0, it has a downward velocity of 0.5m/s as it passes through the point of static equilibrium. Determine the system period, t. k m Answer: Next page vious page P Type here to search
A spring with a spring constant k of 20 pounds per foot is loaded with a 10-pound weight and allowed to reach equilibrium. It is then displaced 1 foot downward and released. If the weight experiences a retarding force in pounds equal to four times the velocity at every point, find the equation of motion. y(t) = where t is time (in seconds) and y(t) is displacement (in feet).
A 3.7kg body is suspended from a spring of constant k = 6.1kN/m. At time t = 0, it has a downward velocity of 0.5m/s as it passes through the point of static equilibrium. Determine the system period, t. k 771 Answer: Next page ous page Type here to search
The 20-kg block is subjected to the action of the harmonic force F = (90 cos 6t) N, where t is in seconds. (Figure 1) Figure k= 400 N/m k = 400 N/m O 20 kg F= 90 cos 6t 125 N.s/m 1 of 1 Part A Write the equation which describes the steady-state motion. Express your answer in terms of t. Express your answer using three significant figures. Express the phase in radians and the final result in meters. x = Submit 17 ΑΣΦ Provide Feedback ↓↑ Request Answer vec ? m Next >
A 5.0 kg weight is hanging from an ideal spring with a k of 250N/m. The block is pulled down .075m from equilibrium. Will the new period for the motion of the block be greater than, less than, or equal to the period for the motion of the block if it were pulled down .15m? Why? The first block is brought to rest at equilibrium again. An identical block is launched up into the first block. The new block is moving with an initial speed of v=5.0m/s when it collides with and sticks to the original block. Calculate the maximum compression of the spring after the collision of the two blocks.
1. An automobile of mass M=5400 kg is moving at a speed of 30 m/s.The engine is disengaged suddenly at t= 0 sec. Assume that the equation of motion after t-0 is given by 5400 v- dv = -8.276 v-2000 dx Where v= v(t) is the speed (m/sec) of the car at t. The left side represents Mv (dv/dx).The first term on the right side is the aerodynamics drag, \\\\\\\\\|\||\\\|||||| || lland the second term is the rolling resistance of the tires. Calculate how far the car travels until the speed reduces to 15 m/sec.(hint: the equation of motion may be integrated as 5400 v dv 30 = Jdx=x 8.276 v+2000 Evaluate the preceding equation using Simpson rule).
A 8.6kg body is suspended from a spring of constant k = 6.9kN/m. At time t = 0, it has a downward velocity of 0.5m/s as it passes through the point of static equilibrium. Determine the natural frequency of the system in rad/s. k m. Answer: Next page evious page ce 9Type here to search
The 20-kg block is subjected to the action of the harmonic force F = (90 cos 6t) N, where t is in seconds. (Figure 1) Figure k-400 N/m Sw 20 kg k=400 N/m 00 F-90 cos 6r c-125 N-s/m < 1 of 1 Part A Write the equation which describes the steady-state motion. Express your answer in terms of t. Express your answer using three significant figures. Express the phase radians and the final result in meters. ΠΙΑΣΦ 4 ? m
A 20-lb weight is suspended from the end of a vertical spring having a spring constant of 40 Ib/ft and is allowed to reach equilibrium. It is then set into motion by stretching the spring 2 in from its equilibrium position and releasing the mass from rest. Find the position of the weight at any time t if there is no external force and no air resistance.
A spring for which k=750 N/m, is suspended from a fixed support. A mass of 30 kg is attached to its lower end and the system is brought to rest. Find the position of the mass at time t=5 sec if a force equal to 20 sin 2t is applied to it.
A5 lbf weight is suspended froma vertical spring having a stiffness k = 50 lbf/ft. An impressed force F = 0.25 sin 8t lbę where t is in seconds is acting on the weight. Determine the equation of motion of the weight when it is pulled down 3 inches from the equilibrium position and released from rest.