Answer:
![x(t) = e^-2t(c_1cos4t+c_2cos4t)](https://tex.z-dn.net/?f=x%28t%29%20%3D%20e%5E-2t%28c_1cos4t%2Bc_2cos4t%29)
![x(t) = \sqrt{5}/2* e^-2tsin(4t+4.24)](https://tex.z-dn.net/?f=x%28t%29%20%3D%20%5Csqrt%7B5%7D%2F2%2A%20e%5E-2tsin%284t%2B4.24%29)
t =1.294
Explanation:
Let m be the mass attached, let k be the spring constant and let p be the positive damping constant. The Newton's second law for the system is
![md^2x/dt^2 = -kx-\beta dx/dt](https://tex.z-dn.net/?f=md%5E2x%2Fdt%5E2%20%3D%20-kx-%5Cbeta%20dx%2Fdt)
where x(t) is the displacement from the equilibrium position. The equation can be transformed into
![d^2x/dt^2 +\beta /m*(dx/dt)+k/m*x=0](https://tex.z-dn.net/?f=d%5E2x%2Fdt%5E2%20%2B%5Cbeta%20%2Fm%2A%28dx%2Fdt%29%2Bk%2Fm%2Ax%3D0)
(a) Let's determine the equation of motion. We must convert units of weight into units of mass
![m=W/g = 3.2/32 =1/10 slug](https://tex.z-dn.net/?f=m%3DW%2Fg%20%3D%203.2%2F32%20%3D1%2F10%20slug)
From Hooke's law we can calculate the spring constant k.
![k = W/s = 2/1 = 2lb/ft](https://tex.z-dn.net/?f=k%20%3D%20W%2Fs%20%3D%202%2F1%20%3D%202lb%2Fft)
If we put m = 1/10 slugs, k = 2 lb/ft and
= 0.4 into the DE, we get
![d^2x/dt^2 +4*(dx/dt)+20*x=0](https://tex.z-dn.net/?f=d%5E2x%2Fdt%5E2%20%2B4%2A%28dx%2Fdt%29%2B20%2Ax%3D0)
The auxiliary equation is m^2+4*m+20 = 0 and its solutions are m_1 = -2-4_i and m_2 = -2 + 4i. The general solution is then
![x(t) = e^-2t(c_1cos4t+c_2cos4t)](https://tex.z-dn.net/?f=x%28t%29%20%3D%20e%5E-2t%28c_1cos4t%2Bc_2cos4t%29)
From the initial conditions
![x(0) = -1ft , x'(0) = 0ft/s](https://tex.z-dn.net/?f=x%280%29%20%3D%20-1ft%20%2C%20x%27%280%29%20%3D%200ft%2Fs)
we can find the equation of motion
![x(t) = e^-2t(-cos4t-1/2sin4t)](https://tex.z-dn.net/?f=x%28t%29%20%3D%20e%5E-2t%28-cos4t-1%2F2sin4t%29)
(b) We need to express the equation of motion in the alternative form
first attachment
The amplitude is
![A=\sqrt{c_1^2+c_2^2}=\sqrt{(-1)^2+(-1/2)^2}= \sqrt{5}/2](https://tex.z-dn.net/?f=A%3D%5Csqrt%7Bc_1%5E2%2Bc_2%5E2%7D%3D%5Csqrt%7B%28-1%29%5E2%2B%28-1%2F2%29%5E2%7D%3D%20%5Csqrt%7B5%7D%2F2)
and the phase angle is
sin∅=![c_1/A=-2/\sqrt{5}](https://tex.z-dn.net/?f=c_1%2FA%3D-2%2F%5Csqrt%7B5%7D)
cos∅=![c_2/A=-1/\sqrt{5}](https://tex.z-dn.net/?f=c_2%2FA%3D-1%2F%5Csqrt%7B5%7D)
tan∅=2
∅=![\pi +tan^-1*2](https://tex.z-dn.net/?f=%5Cpi%20%2Btan%5E-1%2A2)
=4.24
![x(t) = \sqrt{5}/2* e^-2tsin(4t+4.24)](https://tex.z-dn.net/?f=x%28t%29%20%3D%20%5Csqrt%7B5%7D%2F2%2A%20e%5E-2tsin%284t%2B4.24%29)
(c) lb find the wanted time, we will need the derivative of the equation of motion
x'(t)=![5*e^-2tsin4t](https://tex.z-dn.net/?f=5%2Ae%5E-2tsin4t)
The mass passes through the equilibrium when
x(t) =0
![e^-2t(-cos4t-1/2sin4t)=0\\-cos4t-1/2sin4t=0\\tan4t=-2\\4t=-tan^-1+k\pi](https://tex.z-dn.net/?f=e%5E-2t%28-cos4t-1%2F2sin4t%29%3D0%5C%5C-cos4t-1%2F2sin4t%3D0%5C%5Ctan4t%3D-2%5C%5C4t%3D-tan%5E-1%2Bk%5Cpi)
t =-0.2786+k
/4
The velocity at those times is
second attachment
The first (positive) time at which the mass passes through the equilibrium heading upward is for k = 2 (when x' < 0).
t =1.294