Answer to Question #93972 in Classical Mechanics for ayush meena

Question

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)

Answer 1

$ma=m\frac{d^2x}{dt^2}=Kx$

$x=e^{\lambda t}$

We have

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

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

$x(0)=0=c_1 +c_2 \to c_1 =-c_2$

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

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

Thus,

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

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