Question

1) A weight of 18 lbs adheres to a spring and stretches it 4 inches. Find the motion equation if the weight is thrown from the balance position with a downward-directed speed of 2 ft/sec. Find the time when the weight first passes through the equilibrium point upwards.

2) A weight of 24 pounds is attached to a spring whose constant is 5 lb/ft. If the weight is released from a point 3 feet below the balance point in a medium that offers a resistance numerically equal to 3 times the instantaneous speed. Answer the following:

a) What kind of cushioning do we have in this case?

b) Find the motion equation

c) At what point is the weight when t is 1 sec?

3) A mass of 4 kg attached to a spring whose constant is 20 N/m is in a equilibrium position. Starting with t = 0, an external force: f(t) = e^t sin t , is applied to the system. Find the motion equation if the mass moves in a medium that gives you a resistance numerically equal to 8 times the instantaneous velocity. Plot the graph of the motion equation in the range t∈[0 ,10].

4) An L-R-C series circuit contains L = 1/2 H (Henry), R = 4.5 Ω, C = 1/10 F (farad) and E(t) = 110 volts. Determine the instantaneous load, q(t), on the capacitor if q(0)=1 and q'(0) = 0. What's the load after a long time?

5) The torsional movement of a suspended weight at the end of an elastic axis is governed by the differential equation IӨ'' + cӨ' + kӨ' = T(t), where “I” is the moment of inertia, “c” cushioning, “k” the elastic constant of the shaft and T(t) the applied torque. Find Ө (t) if after rest, a torsional force Tt=e-2t is applied to a system where I=1/4, c = 1, k = 4.

Answer 1

1) The mass spring equation of motion

The general solution of this equation is

at t=0, x =0, v = 2ft/sec gives.

now from the velocity equation

to find the spring constant, use the stretch limit

where s = 4 inches mg = 18 lbs weight , we get the constant parameters

So, the motion equation and solution is

time to pass the equilibrium position when x(t) = 0. That means

2) weight W = 24 lbs, Spring constant K = 5 lb/ft , resistance 3 times the velocity

In this case, the equation of motion is

the solution of this equation will have an auxiliary equation

this is producing complex roots with a solution

at t = 0 , x = 3 that gives A1 = 3

This is considered as a damped harmonic motion. if the frequency is divided into real and imaginary then

the velocity is determined by

at t = 1 sec

generally a spiral trajectory