Answer to Question #252247 in Mechanics | Relativity for Anuj

Question #252247

A force of 2 pounds stretches a spring 1 foot. A mass weighing 3.2 pounds is attached to the

spring, and the system is then immersed in a medium that offers a damping force that is

numerically equal to 0.4 times the instantaneous velocity. (a) Find the equation of motion

if the mass is initially released from rest from a point 1 foot above the equilibrium position.

(b) Determine the position and velocity of the mass at t = pi/2 sec? (c) Find the first time at

which the mass passes through the equilibrium position heading upward.



1
Expert's answer
2021-10-18T11:08:40-0400

Solution.

F=kx    k=Fx=21=2;F=kx\implies k=\dfrac{F}{x}=\dfrac{2}{1}=2;

W=mg    3.2=32m    m=0.1slugs.W=mg\implies 3.2=32m\implies m=0.1 slugs.

β=0.4;\beta=0.4;

mx+βx+kx=0;mx^{''}+\beta x^{'}+kx=0;

0.1x+0.4x+2x=0;0.1x^{''}+0.4x^{'}+2x=0;

x+4x+20x=0;x^{''}+4x^{'}+20x=0;

2λ=4;λ=2,ω2=202\lambda=4; \lambda=2, \omega^2=20

x(t)=eλt(C1cos((ω2λ2t)+C2sin(ω2λ2t);x(t)=e^{-\lambda t}(C_1cos(\sqrt({\omega^2-\lambda^2}t)+C_2sin(\sqrt{\omega^2-\lambda^2}t);

x(t)=e2t(C1cos(4t)+C2sin(4t);x(t)=e^{2 t}(C_1cos(4t)+C_2sin(4t);

x(0)=0;x(0)=1;C1=1;x^{'}(0)=0; x(0)=-1; C_1=-1;

x(0)=2C1+4C2=2+4C2    C2=12;x^{'}(0)=-2C_1+4C_2=2+4C_2\implies C_2=-\dfrac{1}{2};

x(t)=e2t(cos(4t)12sin(4t));x(t)=e^{-2t}(-cos(4t)-\dfrac{1}{2}sin(4t));

x(π2)=eπ(cos(2π)12sin(2π))=eπ.x(\dfrac{\pi}{2})=e^{-\pi}(-cos(2\pi)-\dfrac{1}{2}sin(2\pi))=e^{-\pi}.

Answer: a)x(t)=e2t(cos(4t)12sin(4t));a)x(t)=e^{-2t}(-cos(4t)-\dfrac{1}{2}sin(4t));

b)eπ.b)e^{-\pi}.


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