Answer:
[tex]\mathbf{x(t) = \dfrac{ \sqrt 5}{2}e^{-t} }[/tex]
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
SOLUTION:
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;
[tex]C \dfrac{dx}{dt}= F_d[/tex]
[tex]C* 0.25 = 1[/tex]
C = [tex]\dfrac{1 \ N }{0.25 \ m/s}[/tex]
C = 4 Ns/m
NOW;
[tex]m \dfrac{d^2x}{dt^2}+ C \dfrac{dx}{dt}+ kx = 0[/tex]
Divide through by m; we have;
[tex]\dfrac{m}{m}\dfrac{d^2x}{dt^2}+ \dfrac{C}{m} \dfrac{dx}{dt}+ \dfrac{k}{m} x= 0[/tex]
[tex]\dfrac{d^2x}{dt^2}+ \dfrac{C}{m} \dfrac{dx}{dt}+\dfrac{k}{m}x= 0[/tex]
[tex]\dfrac{d^2x}{dt^2}+ \dfrac{4}{2} \dfrac{dx}{dt}+\dfrac{10}{2}x= 0[/tex]
[tex]\dfrac{d^2x}{dt^2}+ 2 \dfrac{dx}{dt}+5x= 0[/tex]
we all know that:
[tex]x(t) = Ae^{(\alpha \ t)}[/tex] ------ (1)
SO;
[tex]\alpha ^2 + 2\alpha + 5 = 0[/tex]
[tex]\alpha = \dfrac{-2 \pm \sqrt{(4)-(4*5)}}{2}[/tex]
[tex]\alpha = -1 \pm 2i[/tex]
Thus ;
[tex]x(t) = e^{-t}[A \sin (2t)+ B cos (2t)][/tex] ------------ (1)
However;
[tex]\dfrac{dx}{dt} = e^{-t}[A \sin (2t)+ B cos (2t)]+ 2e ^{-t} [A \cos (2t)- B \ Sin (2t)][/tex] ------- (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 ;
[tex]x(0) = -1 \ m[/tex]
[tex]\dfrac{dx}{dt}|_{t=0} = 2 \ m/s[/tex]
x(0) ⇒ B = -1
[tex]\dfrac{dx}{dt}|_{t=0} =- B +2A[/tex]
[tex]=- 1 +2A[/tex]
1 = 2A
A = [tex]\dfrac{1}{2}[/tex]
From (1)
[tex]x(t) = e^{-t}[A \sin (2t)+ B cos (2t)][/tex]
[tex]x(t) = e^{-t}[\dfrac{1}{2} \sin (2t)+ (-1) cos (2t)][/tex]
[tex]x(t) = e^{-t}[\dfrac{1}{2} \sin (2t)-cos (2t)][/tex]
Assuming;
[tex]A cos \ \phi = \dfrac{1}{2}[/tex]
[tex]A sin \ \phi = 1[/tex]
Therefore:
[tex]A = \sqrt{\dfrac{1}{4}+1}[/tex]
[tex]A = \sqrt{\dfrac{1+4}{4}}[/tex]
[tex]A = \sqrt{\dfrac{5}{4}}[/tex]
[tex]A ={ \dfrac{ \sqrt5}{ \sqrt4}}[/tex]
[tex]A ={ \dfrac{ \sqrt5}{ 2}}[/tex]
where;
[tex]\phi = tan ^{-1} (2)[/tex]
Therefore;
[tex]x(t) = \dfrac{\sqrt 5}{2}e^{-t} \ sin (2 t - \phi)[/tex]
From above ; the amplitude is ;
[tex]\mathbf{x(t) = \dfrac{ \sqrt 5}{2}e^{-t} }[/tex]
