Edits to “Statement”, “Solution”, “Answer”

Tete edited
revision #20285 parent #20284 ← older newer →
@@ -1,20 +1,20 @@
### Statement
−$3.7.8.$ [Insert the problem 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
−![For problem $3.7.8$ |664x385, 31%](../../img/3.7.8/Savchenko.png)
+![For problem $3.7.8$ |664x385, 60%](../../img/3.7.8/Savchenko.png)
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
−\[P_\perp=(\rho_1\sqrt{(c_1t)^2+(ut)^2}+\rho_2\sqrt{(c_2t)^2+(ut)^2})u.\]
+\[P_\perp=\left(\rho_1\sqrt{(c_1t)^2+(ut)^2}+\rho_2\sqrt{(c_2t)^2+(ut)^2}\right)u.\]
As a result,
−\[F_\perp=P_\perp/t=\rho_1c_1^2\cdot\frac{u}{c_1}\sqrt{1+\left(\frac{u}{c_1}\right)^2}+\rho_2c_2^2\cdot\frac{u}{c_2}\sqrt{1+\left(\frac{u}{c_2}\right)^2}\]
+\[F_\perp=P_\perp/t=\rho_1c_1^2\cdot\frac{u}{c_1}\sqrt{1+\left(\frac{u}{c_1}\right)^2}+\rho_2c_2^2\cdot\frac{u}{c_2}\sqrt{1+\left(\frac{u}{c_2}\right)^2}.\]
−Under the assumption that $F_\parallel\gg F_\perp=\rho_1c_1^2=\rho_2c_2^2$, it follows that $c_1,c_2\gg u$, and we have the approximation
+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
\[F_\perp\approx F_\parallel\left(\frac{u}{c_1}+\frac{u}{c_2}\right).\]
@@ -24,4 +24,4 @@Solution
#### Answer
−[Insert a concise answer or boxed result]
+See the solution.