Edits to “Statement”, “Answer”
en/3.5.41.md
+2 −2
| @@ -1,6 +1,6 @@ | |||
| ### Statement | |||
| − | $3.5.41.$ [Insert the problem statement] | ||
| + | $3.5.41.$ Particles of mass $m$ each fly out of the source at time $t=0$ with almost zero initial velocity. Immediately after they fly out, a force $F=F_0\sin\omega t$ begins to act on them. Determine the velocity of the particles after time $t$ after departure. What is the average velocity of these particles? At what distance from the source is the highest velocity achieved? Answer these questions for the particles emitted at time $t=\pi/(2\omega), t=\pi/\omega$. | ||
| ### 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 $x=-2F_0(n-1)\pi/(m\omega^2)$. | |||
| @@ -38,4 +38,4 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | See the solution. | ||
| @@ -1,6 +1,6 @@ | |||
| ### Statement | ### Statement | ||
| $3.5.41.$ [Insert the problem statement] | $3.5.41.$ Particles of mass $m$ each fly out of the source at time $t=0$ with almost zero initial velocity. Immediately after they fly out, a force $F=F_0\sin\omega t$ begins to act on them. Determine the velocity of the particles after time $t$ after departure. What is the average velocity of these particles? At what distance from the source is the highest velocity achieved? Answer these questions for the particles emitted at time $t=\pi/(2\omega), t=\pi/\omega$. | ||
| ### Solution | ### 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 | 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 | ||
| \[a=\frac{F_0}{m}\sin(\omega t),\] | \[a=\frac{F_0}{m}\sin(\omega t),\] | ||
| \[v=\frac{F_0}{m\omega}[\cos(\omega t_*)-\cos(\omega t)],\] | \[v=\frac{F_0}{m\omega}[\cos(\omega t_*)-\cos(\omega t)],\] | ||
| \[x=\frac{F_0}{m\omega^2}[\omega(t-t_*)\cos(\omega t_*)-\sin(\omega t)+\sin(\omega t_*)].\] | \[x=\frac{F_0}{m\omega^2}[\omega(t-t_*)\cos(\omega t_*)-\sin(\omega t)+\sin(\omega t_*)].\] | ||
| When $t_*=0$, we have | When $t_*=0$, we have | ||
| \[v=\frac{F_0}{m\omega}[1-\cos(\omega t)],\] | \[v=\frac{F_0}{m\omega}[1-\cos(\omega t)],\] | ||
| \[x=\frac{F_0}{m\omega^2}[\omega t-\sin(\omega t)].\] | \[x=\frac{F_0}{m\omega^2}[\omega t-\sin(\omega t)].\] | ||
| 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. | 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 | When $t_*=\pi/(2\omega)$, we have | ||
| \[v=-\frac{F_0}{m\omega}\cos(\omega t),\] | \[v=-\frac{F_0}{m\omega}\cos(\omega t),\] | ||
| \[x=\frac{F_0}{m\omega^2}[1-\sin(\omega t)].\] | \[x=\frac{F_0}{m\omega^2}[1-\sin(\omega t)].\] | ||
| Thus, the average velocity is $0$, and the maximum velocity is $F_0/(m\omega)$ achieved when $t=(2n-1)\pi/\omega$ and hence $x=F_0/(m\omega^2)$. | 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 | When $t_*=\pi/\omega$, we have | ||
| \[v=-\frac{F_0}{m\omega}[1+\cos(\omega t)],\] | \[v=-\frac{F_0}{m\omega}[1+\cos(\omega t)],\] | ||
| \[x=-\frac{F_0}{m\omega^2}[\omega(t-\pi/\omega)+\sin(\omega t)].\] | \[x=-\frac{F_0}{m\omega^2}[\omega(t-\pi/\omega)+\sin(\omega t)].\] | ||
| Thus, the average velocity is $-F_0/(m\omega)$, and the maximum velocity is $0$ achieved when $t=(2n-1)\pi/\omega$ and hence $x=-2F_0(n-1)\pi/(m\omega^2)$. | Thus, the average velocity is $-F_0/(m\omega)$, and the maximum velocity is $0$ achieved when $t=(2n-1)\pi/\omega$ and hence $x=-2F_0(n-1)\pi/(m\omega^2)$. | ||
| @@ -38,4 +38,4 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | See the solution. | ||