Answer to Question #93972 in Classical Mechanics for ayush meena

Question #93972
A particle of mass m is subject to a force f(X)= Kx. the initial position is 0 and the initial speed is V. Find x(t)
1
Expert's answer
2019-09-09T11:06:47-0400
ma=md2xdt2=Kxma=m\frac{d^2x}{dt^2}=Kx

x=eλtx=e^{\lambda t}

We have


λ2=Kmλ1=Km,λ2=Km\lambda^2=\frac{K}{m}\to \lambda_1=\sqrt{\frac{K}{m}},\lambda_2=-\sqrt{\frac{K}{m}}

x(t)=c1exp(Kmt)+c2exp(Kmt)x(t)=c_1 \exp{\left(\sqrt{\frac{K}{m}}t\right)}+c_2 \exp{\left(-\sqrt{\frac{K}{m}}t\right)}

x(0)=0=c1+c2c1=c2x(0)=0=c_1 +c_2 \to c_1 =-c_2

v(t)=c1Km(exp(Kmt)+exp(Kmt))v(t)=c_1 \sqrt{\frac{K}{m}}\left(\exp{\left(\sqrt{\frac{K}{m}}t\right)}+\exp{\left(-\sqrt{\frac{K}{m}}t\right)}\right)

v(0)=v=c1Km(1+1)v(0)=v=c_1 \sqrt{\frac{K}{m}}\left(1+1\right)

Thus,


x(t)=0.5vmK(exp(Kmt)exp(Kmt))x(t)=0.5v\sqrt{\frac{m}{K}}\left(\exp{\left(\sqrt{\frac{K}{m}}t\right)}-\exp{\left(-\sqrt{\frac{K}{m}}t\right)}\right)

x(t)=vmKsh(Kmt)x(t)=v\sqrt{\frac{m}{K}}\sh{\left(\sqrt{\frac{K}{m}}t\right)}


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