Edit to “Solution”
en/3.7.8.md
+5 −7
| ### Statement | |||
| $3.7.8.$ A string consisting of two parts with linear densities $\rho_1$ and $\rho_2$ is stretched by longitudinal forces $F_\parallel$. At the point where the parts are connected, the string is pulled by a transverse force $F_\perp$. How does the shape of the string change with time? | |||
| ### Solution | |||
|  | |||
| @@ -8,20 +8,18 @@Solution | |||
| The speeds of propagation of the waveform to the left and the right are $c_1=\sqrt{F_\parallel/\rho_1}$ and $c_2=\sqrt{F_\parallel/\rho_2}$, respectively. Consequently, the mass of the portion of the rope that has transverse momentum increases linearly with time, and its transverse speed $u$ must be constant, because the transverse force is constant. After time $t$, transverse momentum of the waveform is | |||
| − | \[P_\perp= | ||
| + | \[P_\perp=(\rho_1c_1t+\rho_2c_2t)u.\] | ||
| + | Note that even though the lengths of the stretched parts of the rope are $\sqrt{(c_1t)^2+(ut)^2}$ and $\sqrt{(c_2t)^2+(ut)^2}$, they used to have lengths $c_1t$ and $c_2t$, so they must have masses $\rho_1c_1t$ and $\rho_2c_2t$, respectively. | ||
| + | |||
| As a result, | |||
| − | \[F_\perp=P_\perp/t=\rho_1c_1^2\cdot\frac{u}{c_1} | ||
| + | \[F_\perp=P_\perp/t=\rho_1c_1^2\cdot\frac{u}{c_1}+\rho_2c_2^2\cdot\frac{u}{c_2}=F_\parallel\left(\frac{u}{c_1}+\frac{u}{c_2}\right).\] | ||
| − | Under the assumption that $F_\perp\ll F_\parallel=\rho_1c_1^2=\rho_2c_2^2$, it follows that $u\ll c_1,c_2$, and we have the approximation | ||
| − | |||
| − | \[F_\perp\approx F_\parallel\left(\frac{u}{c_1}+\frac{u}{c_2}\right).\] | ||
| − | |||
| Therefore, | |||
| − | \[u | ||
| + | \[u=\frac{c_1c_2}{c_1+c_2}\cdot\frac{F_\perp}{F_\parallel}.\] | ||
| #### Answer | |||
| See the solution. | |||
| ### Statement | ### Statement | ||
| $3.7.8.$ A string consisting of two parts with linear densities $\rho_1$ and $\rho_2$ is stretched by longitudinal forces $F_\parallel$. At the point where the parts are connected, the string is pulled by a transverse force $F_\perp$. How does the shape of the string change with time? | $3.7.8.$ A string consisting of two parts with linear densities $\rho_1$ and $\rho_2$ is stretched by longitudinal forces $F_\parallel$. At the point where the parts are connected, the string is pulled by a transverse force $F_\perp$. How does the shape of the string change with time? | ||
| ### Solution | ### Solution | ||
|  |  | ||
| @@ -8,20 +8,18 @@Solution | |||
| The speeds of propagation of the waveform to the left and the right are $c_1=\sqrt{F_\parallel/\rho_1}$ and $c_2=\sqrt{F_\parallel/\rho_2}$, respectively. Consequently, the mass of the portion of the rope that has transverse momentum increases linearly with time, and its transverse speed $u$ must be constant, because the transverse force is constant. After time $t$, transverse momentum of the waveform is | The speeds of propagation of the waveform to the left and the right are $c_1=\sqrt{F_\parallel/\rho_1}$ and $c_2=\sqrt{F_\parallel/\rho_2}$, respectively. Consequently, the mass of the portion of the rope that has transverse momentum increases linearly with time, and its transverse speed $u$ must be constant, because the transverse force is constant. After time $t$, transverse momentum of the waveform is | ||
| \[P_\perp= |
\[P_\perp=(\rho_1c_1t+\rho_2c_2t)u.\] | ||
| Note that even though the lengths of the stretched parts of the rope are $\sqrt{(c_1t)^2+(ut)^2}$ and $\sqrt{(c_2t)^2+(ut)^2}$, they used to have lengths $c_1t$ and $c_2t$, so they must have masses $\rho_1c_1t$ and $\rho_2c_2t$, respectively. | |||
| As a result, | As a result, | ||
| \[F_\perp=P_\perp/t=\rho_1c_1^2\cdot\frac{u}{c_1} |
\[F_\perp=P_\perp/t=\rho_1c_1^2\cdot\frac{u}{c_1}+\rho_2c_2^2\cdot\frac{u}{c_2}=F_\parallel\left(\frac{u}{c_1}+\frac{u}{c_2}\right).\] | ||
| Under the assumption that $F_\perp\ll F_\parallel=\rho_1c_1^2=\rho_2c_2^2$, it follows that $u\ll c_1,c_2$, and we have the approximation | |||
| \[F_\perp\approx F_\parallel\left(\frac{u}{c_1}+\frac{u}{c_2}\right).\] | |||
| Therefore, | Therefore, | ||
| \[u |
\[u=\frac{c_1c_2}{c_1+c_2}\cdot\frac{F_\perp}{F_\parallel}.\] | ||
| #### Answer | #### Answer | ||
| See the solution. | See the solution. | ||