Respuesta :
Answer:
[tex]x(t)=-5sint2t[/tex]
Step-by-step explanation:
Data given
Force=720N
displacement=4m
mass = 45kg
velocity=8m/s
We can determine the spring constant using hooks law
F=ke
K=f/e
K=720/4=180N/M
also w can write newton second law of motion for the system
[tex]\frac{d^2x}{dt^2}+\frac{k}{m}x =0[/tex]
if we substitute the values of the mass and spring constant we have
[tex]\frac{d^2x}{dt^2}+\frac{k}{m}x =0\\\frac{d^2x}{dt^2}+4x=0\\[/tex]
The solution of the differential equation is of the form
[tex]\frac{d^2x}{dt^2}+w^2x=0 \\ x(t)=C_1coswt+C_2sinwt\\w^2=4, \\w=2\\ x(t)=C_1cos2t+C_2sin2t\\[/tex]
Next we solve for the initial and final solution we arrive at
[tex]x(0)=0m\\\\x^{'} (0)=10m/s[/tex]
Hence the final solution is of the form
[tex]x(t)=-5sint2t[/tex]
The final equation of motion of the mass and spring system given in the question is;
x(t) = -4sin 2t
We are given;
Force; F = 720N
displacement; s = 4m
mass; m = 45kg
velocity; v = 8m/s
- From newton's second law, we can have the expression for this system as;
d²x/dt² + (k/m)x = 0
From hooke's law, we know that;
F = ks
where k is spring constant
k = F/s
k = 720/4
k = 180 N/m
Thus;
d²x/dt² + (720/180)x = 0
d²x/dt² + 4x = 0
The general form of this is;
x'' + ω²x = 0
Thus;
ω² = 4
ω = 2
We know that the solution to it is given by the general expression;
x(t) = c₁ cos ωt + c₂ sinωt
Thus;
x(t) = c₁ cos 2t + c₂ sin2t
When we solve at the initial conditions of the equation, we arrive at;
x(0) = 0 and x'(0) = -8 m/s
Thus, the equation is; x(t) = -4sin 2t
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