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,

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

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