Respuesta :
Given that,
Weight = 4 pound
[tex]W=4\ lb[/tex]
Stretch = 2 feet
Let the force be F.
The elongation of the spring after the mass attached is
[tex]x=2-1=1\ feet[/tex]
(a). We need to calculate the value of spring constant
Using Hooke's law
[tex]F=kx[/tex]
[tex]k=\dfrac{F}{x}[/tex]
Where, F = force
k = spring constant
x = elongation
Put the value into the formula
[tex]k=\dfrac{4}{1}[/tex]
[tex]k=4[/tex]
(b). We need to calculate the mass
Using the formula
[tex]F=mg[/tex]
[tex]m=\dfrac{F}{g}[/tex]
Where, F = force
g = acceleration due to gravity
Put the value into the formula
[tex]m=\dfrac{4}{32}[/tex]
[tex]m=\dfrac{1}{8}\ lb[/tex]
We need to calculate the natural frequency
Using formula of natural frequency
[tex]\omega=\sqrt{\dfrac{k}{m}}[/tex]
Where, k = spring constant
m = mass
Put the value into the formula
[tex]\omega=\sqrt{\dfrac{4}{\dfrac{1}{8}}}[/tex]
[tex]\omega=\sqrt{32}[/tex]
[tex]\omega=4\sqrt{2}[/tex]
(c). We need to write the differential equation
Using differential equation
[tex]m\dfrac{d^2x}{dt^2}+kx=0[/tex]
Put the value in the equation
[tex]\dfrac{1}{8}\dfrac{d^2x}{dt^2}+4x=0[/tex]
[tex]\dfrac{d^2x}{dt^2}+32x=0[/tex]
(d). We need to find the solution for the position
Using auxiliary equation
[tex]m^2+32=0[/tex]
[tex]m=\pm i\sqrt{32}[/tex]
We know that,
The general equation is
[tex]x(t)=A\cos(\sqrt{32t})+B\sin(\sqrt{32t})[/tex]
Using initial conditions
(I). [tex]x(0)=2[/tex]
Then, [tex]x(0)=A\cos(\sqrt{32\times0})+B\sin(\sqrt{32\times0})[/tex]
Put the value in equation
[tex]2=A+0[/tex]
[tex]A=2[/tex].....(I)
Now, on differentiating of general equation
[tex]x'(t)=-\sqrt{32}A\sin(\sqrt{32t})+\sqrt{32}B\cos(\sqrt{32t})[/tex]
Using condition
(II). [tex]x'(0)=0[/tex]
Then, [tex]x'(0)=-\sqrt{32}A\sin(\sqrt{32\times0})+\sqrt{32}B\cos(\sqrt{32\times0})[/tex]
Put the value in the equation
[tex]0=0+\sqrt{32}B[/tex]
So, B = 0
Now, put the value in general equation from equation (I) and (II)
So, The general solution is
[tex] x(t)=2\cos\sqrt{32t}[/tex]
(e). We need to calculate the time
Using formula of time
[tex]T=\dfrac{2\pi}{\omega}[/tex]
Put the value into the formula
[tex]T=\dfrac{2\pi}{4\sqrt{2}}[/tex]
[tex]T=1.11\ sec[/tex]
Hence, (a). The value of spring constant is 4.
(b). The natural frequency is 4√2.
(c). The differential equation is [tex]\dfrac{d^2x}{dt^2}+32x=0[/tex]
(d). The solution for the position is [tex] x(t)=2\cos\sqrt{32t}[/tex]
(e). The time period is 1.11 sec.
