Problem 3.5.41^∗

Solution

From $F=F_0\sin(\omega t)$ and the initial conditions $v=0$ and $x=0$ at $t=t_*$ (depending on when the particles are emitted), we have

$$a=\frac{F_0}{m}\sin(\omega t),$$

$$v=\frac{F_0}{m\omega}[\cos(\omega t_*)-\cos(\omega t)],$$

$$x=\frac{F_0}{m\omega^2}[\omega(t-t_*)\cos(\omega t_*)-\sin(\omega t)+\sin(\omega t_*)].$$

When $t_*=0$, we have

$$v=\frac{F_0}{m\omega}[1-\cos(\omega t)],$$

$$x=\frac{F_0}{m\omega^2}[\omega t-\sin(\omega t)].$$

Thus, the average velocity is $F_0/(m\omega)$, and the maximum velocity is $2F_0/(m\omega)$ achieved when $t=(2n-1)\pi/\omega$ and hence $x=F_0(2n-1)\pi/(m\omega^2)$, where $n$ is a positive integer.

When $t_*=\pi/(2\omega)$, we have

$$v=-\frac{F_0}{m\omega}\cos(\omega t),$$

$$x=\frac{F_0}{m\omega^2}[1-\sin(\omega t)].$$

Thus, the average velocity is $0$, and the maximum velocity is $F_0/(m\omega)$ achieved when $t=(2n-1)\pi/\omega$ and hence $x=F_0/(m\omega^2)$.

When $t_*=\pi/\omega$, we have

$$v=-\frac{F_0}{m\omega}[1+\cos(\omega t)],$$

$$x=-\frac{F_0}{m\omega^2}[\omega(t-\pi/\omega)+\sin(\omega t)].$$

Thus, the average velocity is $-F_0/(m\omega)$, and the maximum velocity is $0$ achieved when $t=(2n-1)\pi/\omega$ and hence $x=-2F_0(n-1)\pi/(m\omega^2)$.