Answer:
the required value is ![x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}](https://tex.z-dn.net/?f=x%28t%29%20%3D%20%5Cfrac%7B7%7D%7B4%7D%20e%5E%7B-3t%7D-%5Cfrac%7B3%7D%7B4%7D%20e%5E%7B-7t%7D)
Explanation:
Given that,
mass, m = 1kg
spring constant k = 21N/M
damping force = ![-\beta\frac{dx}{dt} = \frac{-10dx}{dt}](https://tex.z-dn.net/?f=-%5Cbeta%5Cfrac%7Bdx%7D%7Bdt%7D%20%3D%20%5Cfrac%7B-10dx%7D%7Bdt%7D)
![\beta = 10](https://tex.z-dn.net/?f=%5Cbeta%20%3D%2010)
By Newtons second law ,
The diffrential equation of motion with damping is given by
![m\frac{d^2x}{dt^2} = -kx-\beta\frac{dx}{dt}](https://tex.z-dn.net/?f=m%5Cfrac%7Bd%5E2x%7D%7Bdt%5E2%7D%20%3D%20-kx-%5Cbeta%5Cfrac%7Bdx%7D%7Bdt%7D)
substitute the value of m =1kg, k = 21N/M, and ![\beta = 10](https://tex.z-dn.net/?f=%5Cbeta%20%3D%2010)
![1\frac{d^2x}{dt^2} = -21x=10\frac{dx}{dt}](https://tex.z-dn.net/?f=1%5Cfrac%7Bd%5E2x%7D%7Bdt%5E2%7D%20%3D%20-21x%3D10%5Cfrac%7Bdx%7D%7Bdt%7D)
![\frac{d^2x}{dt^2} + 10\frac{dx}{dt} + 21x = 0](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5E2x%7D%7Bdt%5E2%7D%20%2B%2010%5Cfrac%7Bdx%7D%7Bdt%7D%20%2B%2021x%20%3D%200)
suppose the equation of the form
,
and the auxilliary equation is given by
![m^2 + 10m + 21 = 0\\\\m^2 + 7m+3m+21=0\\\\m(m+7)+3(m+7)=0\\\\(m+7)(m+3)=0\\\\m=-7\\m=-3](https://tex.z-dn.net/?f=m%5E2%20%2B%2010m%20%2B%2021%20%3D%200%5C%5C%5C%5Cm%5E2%20%2B%207m%2B3m%2B21%3D0%5C%5C%5C%5Cm%28m%2B7%29%2B3%28m%2B7%29%3D0%5C%5C%5C%5C%28m%2B7%29%28m%2B3%29%3D0%5C%5C%5C%5Cm%3D-7%5C%5Cm%3D-3)
The general solution for the above differential equation is
![x(t) =C_1e^{-3t}+C_2e^{-7t}](https://tex.z-dn.net/?f=x%28t%29%20%3DC_1e%5E%7B-3t%7D%2BC_2e%5E%7B-7t%7D)
Derivate with respect to t
![x'(t)=-3C_1e^{-3t}-7C_2e^{-7t}](https://tex.z-dn.net/?f=x%27%28t%29%3D-3C_1e%5E%7B-3t%7D-7C_2e%5E%7B-7t%7D)
(a)
since time is 0 then mass is one meter below
so x(0) = 1
Also it start from rest , that implies , velocity is 0 and time is 0
![x'(0) = 0](https://tex.z-dn.net/?f=x%27%280%29%20%3D%200)
substitute the initial condition
![C_1 +C_2 = 1](https://tex.z-dn.net/?f=C_1%20%2BC_2%20%3D%201)
![-3C_1-7C_2=0](https://tex.z-dn.net/?f=-3C_1-7C_2%3D0)
Solve the above equation to get C₁ and C₂
![C_1 =\frac{7}{4} and C_2 = -\frac{3}{4}](https://tex.z-dn.net/?f=C_1%20%3D%5Cfrac%7B7%7D%7B4%7D%20and%20C_2%20%3D%20-%5Cfrac%7B3%7D%7B4%7D)
substitute for C₁ and C₂ in general solution
![x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}](https://tex.z-dn.net/?f=x%28t%29%20%3D%20%5Cfrac%7B7%7D%7B4%7D%20e%5E%7B-3t%7D-%5Cfrac%7B3%7D%7B4%7D%20e%5E%7B-7t%7D)
Thus the required value is ![x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}](https://tex.z-dn.net/?f=x%28t%29%20%3D%20%5Cfrac%7B7%7D%7B4%7D%20e%5E%7B-3t%7D-%5Cfrac%7B3%7D%7B4%7D%20e%5E%7B-7t%7D)