Answer:
![\mathbf{x(t) = \dfrac{ \sqrt 5}{2}e^{-t} }](https://tex.z-dn.net/?f=%5Cmathbf%7Bx%28t%29%20%3D%20%5Cdfrac%7B%20%5Csqrt%205%7D%7B2%7De%5E%7B-t%7D%20%7D)
Step-by-step explanation:
Given that:
mass of the object = 2 kg
A force of 5nt is applied to move the object 0.5m from its equilibrium position.
i.e
Force = 5 newton
Stretchin (x) =0.5 m
Damping force = 1 newton
Velocity = 0.25 m/second
The object is pulled to the left until the spring is stretched lm and then released with the initial velocity of 2m/second to the right
<u>SOLUTION:</u>
If F = kx
Then :
5 N = k(0.5 m)
where ;
k = spring constant.
k = 5 N/0.5 m
k = 10 N/m
the damping force of the object sliding on the table is 1 newton when the velocity is 0.25m/second.
SO;
![C \dfrac{dx}{dt}= F_d](https://tex.z-dn.net/?f=C%20%5Cdfrac%7Bdx%7D%7Bdt%7D%3D%20F_d)
![C* 0.25 = 1](https://tex.z-dn.net/?f=C%2A%200.25%20%3D%201)
C = ![\dfrac{1 \ N }{0.25 \ m/s}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%20%20%5C%20N%20%7D%7B0.25%20%20%5C%20m%2Fs%7D)
C = 4 Ns/m
NOW;
![m \dfrac{d^2x}{dt^2}+ C \dfrac{dx}{dt}+ kx = 0](https://tex.z-dn.net/?f=m%20%5Cdfrac%7Bd%5E2x%7D%7Bdt%5E2%7D%2B%20C%20%5Cdfrac%7Bdx%7D%7Bdt%7D%2B%20kx%20%3D%200)
Divide through by m; we have;
![\dfrac{m}{m}\dfrac{d^2x}{dt^2}+ \dfrac{C}{m} \dfrac{dx}{dt}+ \dfrac{k}{m} x= 0](https://tex.z-dn.net/?f=%5Cdfrac%7Bm%7D%7Bm%7D%5Cdfrac%7Bd%5E2x%7D%7Bdt%5E2%7D%2B%20%5Cdfrac%7BC%7D%7Bm%7D%20%5Cdfrac%7Bdx%7D%7Bdt%7D%2B%20%5Cdfrac%7Bk%7D%7Bm%7D%20x%3D%200)
![\dfrac{d^2x}{dt^2}+ \dfrac{C}{m} \dfrac{dx}{dt}+\dfrac{k}{m}x= 0](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%5E2x%7D%7Bdt%5E2%7D%2B%20%5Cdfrac%7BC%7D%7Bm%7D%20%5Cdfrac%7Bdx%7D%7Bdt%7D%2B%5Cdfrac%7Bk%7D%7Bm%7Dx%3D%200)
![\dfrac{d^2x}{dt^2}+ \dfrac{4}{2} \dfrac{dx}{dt}+\dfrac{10}{2}x= 0](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%5E2x%7D%7Bdt%5E2%7D%2B%20%5Cdfrac%7B4%7D%7B2%7D%20%5Cdfrac%7Bdx%7D%7Bdt%7D%2B%5Cdfrac%7B10%7D%7B2%7Dx%3D%200)
![\dfrac{d^2x}{dt^2}+ 2 \dfrac{dx}{dt}+5x= 0](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%5E2x%7D%7Bdt%5E2%7D%2B%202%20%5Cdfrac%7Bdx%7D%7Bdt%7D%2B5x%3D%200)
we all know that:
------ (1)
SO;
![\alpha ^2 + 2\alpha + 5 = 0](https://tex.z-dn.net/?f=%5Calpha%20%5E2%20%2B%202%5Calpha%20%2B%205%20%3D%200)
![\alpha = \dfrac{-2 \pm \sqrt{(4)-(4*5)}}{2}](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cdfrac%7B-2%20%5Cpm%20%5Csqrt%7B%284%29-%284%2A5%29%7D%7D%7B2%7D)
![\alpha = -1 \pm 2i](https://tex.z-dn.net/?f=%5Calpha%20%3D%20-1%20%5Cpm%202i)
Thus ;
------------ (1)
However;
------- (2)
From the question ; we are being told that;
The object is pulled to the left until the spring is stretched 1 m and then released with the initial velocity of 2m/second to the right.
So ;
![x(0) = -1 \ m](https://tex.z-dn.net/?f=x%280%29%20%3D%20-1%20%5C%20m)
![\dfrac{dx}{dt}|_{t=0} = 2 \ m/s](https://tex.z-dn.net/?f=%5Cdfrac%7Bdx%7D%7Bdt%7D%7C_%7Bt%3D0%7D%20%20%3D%202%20%5C%20m%2Fs)
x(0) ⇒ B = -1
![\dfrac{dx}{dt}|_{t=0} =- B +2A](https://tex.z-dn.net/?f=%5Cdfrac%7Bdx%7D%7Bdt%7D%7C_%7Bt%3D0%7D%20%20%3D-%20B%20%2B2A)
![=- 1 +2A](https://tex.z-dn.net/?f=%3D-%201%20%2B2A)
1 = 2A
A = ![\dfrac{1}{2}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7D)
From (1)
![x(t) = e^{-t}[\dfrac{1}{2} \sin (2t)+ (-1) cos (2t)]](https://tex.z-dn.net/?f=x%28t%29%20%3D%20e%5E%7B-t%7D%5B%5Cdfrac%7B1%7D%7B2%7D%20%5Csin%20%282t%29%2B%20%28-1%29%20cos%20%282t%29%5D)
![x(t) = e^{-t}[\dfrac{1}{2} \sin (2t)-cos (2t)]](https://tex.z-dn.net/?f=x%28t%29%20%3D%20e%5E%7B-t%7D%5B%5Cdfrac%7B1%7D%7B2%7D%20%5Csin%20%282t%29-cos%20%282t%29%5D)
Assuming;
![A cos \ \phi = \dfrac{1}{2}](https://tex.z-dn.net/?f=A%20cos%20%20%20%5C%20%20%5Cphi%20%3D%20%5Cdfrac%7B1%7D%7B2%7D)
![A sin \ \phi = 1](https://tex.z-dn.net/?f=A%20sin%20%5C%20%20%5Cphi%20%3D%201)
Therefore:
![A = \sqrt{\dfrac{1}{4}+1}](https://tex.z-dn.net/?f=A%20%3D%20%5Csqrt%7B%5Cdfrac%7B1%7D%7B4%7D%2B1%7D)
![A = \sqrt{\dfrac{1+4}{4}}](https://tex.z-dn.net/?f=A%20%3D%20%5Csqrt%7B%5Cdfrac%7B1%2B4%7D%7B4%7D%7D)
![A = \sqrt{\dfrac{5}{4}}](https://tex.z-dn.net/?f=A%20%3D%20%5Csqrt%7B%5Cdfrac%7B5%7D%7B4%7D%7D)
![A ={ \dfrac{ \sqrt5}{ \sqrt4}}](https://tex.z-dn.net/?f=A%20%3D%7B%20%5Cdfrac%7B%20%20%5Csqrt5%7D%7B%20%20%5Csqrt4%7D%7D)
![A ={ \dfrac{ \sqrt5}{ 2}}](https://tex.z-dn.net/?f=A%20%3D%7B%20%5Cdfrac%7B%20%20%5Csqrt5%7D%7B%20%202%7D%7D)
where;
![\phi = tan ^{-1} (2)](https://tex.z-dn.net/?f=%5Cphi%20%3D%20tan%20%5E%7B-1%7D%20%282%29)
Therefore;
![x(t) = \dfrac{\sqrt 5}{2}e^{-t} \ sin (2 t - \phi)](https://tex.z-dn.net/?f=x%28t%29%20%3D%20%5Cdfrac%7B%5Csqrt%205%7D%7B2%7De%5E%7B-t%7D%20%5C%20sin%20%282%20t%20-%20%5Cphi%29)
From above ; the amplitude is ;
![\mathbf{x(t) = \dfrac{ \sqrt 5}{2}e^{-t} }](https://tex.z-dn.net/?f=%5Cmathbf%7Bx%28t%29%20%3D%20%5Cdfrac%7B%20%5Csqrt%205%7D%7B2%7De%5E%7B-t%7D%20%7D)
Answer:
its A i just took the test
Answer:
![60\sqrt{3 \\ }](https://tex.z-dn.net/?f=%2060%5Csqrt%7B3%20%5C%5C%20%7D%20)
Step-by-step explanation:
in a cube, the diagonal is the length of a side times square root of 3
Answer:
n=3p
Isolate the variable by dividing each side by factors that don't contain the variable.