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Sagot :
Answer:
The equation of motion is [tex]x(t)=-[/tex][tex]\frac{1}{3} cos4\sqrt{6t}[/tex]
Explanation:
Lets calculate
The weight attached to the spring is 24 pounds
Acceleration due to gravity is [tex]32ft/s^2[/tex]
Assume x , is spring stretched length is ,4 inches
Converting the length inches into feet [tex]x=\frac{4}{12} =\frac{1}{3}feet[/tex]
The weight (W=mg) is balanced by restoring force ks at equilibrium position
mg=kx
[tex]W=kx[/tex] ⇒ [tex]k=\frac{W}{x}[/tex]
The spring constant , [tex]k=\frac{24}{1/3}[/tex]
= 72
If the mass is displaced from its equilibrium position by an amount x, then the differential equation is
[tex]m\frac{d^2x}{dt} +kx=0[/tex]
[tex]\frac{3}{4} \frac{d^2x}{dt} +72x=0[/tex]
[tex]\frac{d^2x}{dt} +96x=0[/tex]
Auxiliary equation is, [tex]m^2+96=0[/tex]
[tex]m=\sqrt{-96}[/tex]
=[tex]\frac{+}{} i4\sqrt{6}[/tex]
Thus , the solution is [tex]x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}[/tex]
[tex]x'(t)=-4\sqrt{6c_1} sin4\sqrt{6t}+c_2[/tex] [tex]4\sqrt{6}[/tex] [tex]cos4\sqrt{6t}[/tex]
The mass is released from the rest x'(0) = 0
[tex]=-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2[/tex] [tex]4\sqrt{6}[/tex] [tex]cos4\sqrt{6(0)}[/tex] =0
[tex]c_2[/tex] [tex]4\sqrt{6} =0[/tex]
[tex]c_2=0[/tex]
Therefore , [tex]x(t)=c_1[/tex] [tex]cos 4\sqrt{6t}[/tex]
Since , the mass is released from the rest from 4 inches
[tex]x(0)= -4[/tex] inches
[tex]c_1 cos 4\sqrt{6(0)} =-\frac{4}{12}[/tex] feet
[tex]c_1=-\frac{1}{3}[/tex] feet
Therefore , the equation of motion is [tex]-\frac{1}{3} cos4\sqrt{6t}[/tex]
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