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Alexphysics edited Jun 15, 2026, 6:26 AM

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			### Statement

			$3.7.6.$
			A longitudinal force F acts on the end of a resting semi-infinite rod for time
			τ. Find the velocity of the rod particles and its deformation in the area of the
			resulting wave if the cross-section of the rod is S, the Young’s modulus of its
			material is E, and the density is ρ. What is the density of the rod in the wave
			region? Find the momentum and energy of the displaced particles of the rod
			after time 0.5τ and 1.5τ from the beginning of the force.



			### Solution


			The longitudinal waves in a thin bar propagate with speed


			$\boxed{c = \sqrt{\frac{E}{\rho}}}$



			and this speed does not depend on the sign of the force.


			The compressive force produces a negative normal stress (compression):


			$\sigma = -\frac{F}{S}$


			By Hooke's law the unit strain is

			$\varepsilon = \sigma / E$:


			$\boxed{\varepsilon = -\frac{F}{SE}}$


			The quantity $\|\varepsilon\|$
			is the relative shortening.
			and the negative sign indicates compression.


			In a progressive plane elastic wave traveling in the positive direction,

			the relationship between the stress $\sigma $ and the particle velocity v is:

			$\sigma = -\rho c v$


			(the minus sign appears because in compression the material moves in the direction of wave propagation). Solving for v:


			$v = -\frac{\sigma}{\rho c} = -\frac{(-F/S)}{\rho c} = \frac{F}{S\rho c}$


			Substituting $c = \sqrt{E/\rho}$:


			$\boxed{v = \frac{F}{S\sqrt{\rho E}}}$


			The particles move in the direction of wave propagation, with constant speed as long as the force is applied.

			Conservation of mass implies that the density \rho' in the deformed region satisfies $\rho' = \rho / (1+\varepsilon)$.

			In the linear regime $(\|\varepsilon\| \ll 1)$ we can approximate:

			$\rho' \approx \rho(1 - \varepsilon) = \rho\left(1 + \frac{F}{SE}\right)$

			Since $\varepsilon $ is negative, the density increases in the compressed region.

			$\boxed{\rho' = \rho\left[1 + \frac{F}{SE}\right]}$

			While the force is applied$ (0 < t \le \tau)$,
			the wave front advances a distance $l(t) = ct$.
			The total mass set into motion up to that instant is:

			$m(t) = \rho S l(t) = \rho S c t%

			Momentum and energy at$ t = 0.5\tau$

			At this instant the force is still acting;
			the disturbance has not yet ceased.

			Mass in motion: $m = \rho S c \cdot 0.5\tau.$
			Impulse:$ p = m v$.

			$p = (\rho S c \cdot 0.5\tau) \left( \frac{F}{S\sqrt{\rho E}} \right) = 0.5 F\tau \frac{\rho c}{\sqrt{\rho E}}$

			Since
			$\rho c = \rho\sqrt{E/\rho} = \sqrt{\rho E}$,

			this simplifies to

			$\boxed{p = 0.5 F\tau}$

			Kinetic energy:
			$K = \frac{1}{2} m v^2$.

			$K = \frac{1}{2} (0.5 \rho S c \tau) \left( \frac{F}{S\sqrt{\rho E}} \right)^2 = \frac{F^2 c \tau}{4 S E}$

			Using $c/E = 1/\sqrt{\rho E}$:

			$K = \frac{F^2 \tau}{4 S \sqrt{\rho E}}$

			$Elastic potential energy: energy density u_e = \frac{1}{2} \sigma \varepsilon = \frac{F^2}{2 E S^2}$.

			Perturbed volume
			$V = S l = 0.5 S c \tau$.

			$U = u_e V = \frac{F^2}{2 E S^2} \cdot 0.5 S c \tau = \frac{F^2 c \tau}{4 E S} = \frac{F^2 \tau}{4 S \sqrt{\rho E}}$

			Total energy:

			$\boxed{W = K + U = \frac{F^2 \tau}{2 S \sqrt{\rho E}}}$

			Momentum and energy at $t = 1.5\tau$

			The force ceased at $t = \tau$. Now the pulse has detached from the end and travels freely with fixed length $L = c\tau$. The total mass contained in the pulse is:

			$m = \rho S L = \rho S c \tau$

			Impulse:$ p' = m v$.

			$ p' = (\rho S c \tau) \left( \frac{F}{S\sqrt{\rho E}} \right) = F\tau \frac{\rho c}{\sqrt{\rho E}} = F\tau$

			$\boxed{p' = F\tau}$

			Kinetic energy:

			$K = \frac{1}{2} m v^2 = \frac{1}{2} (\rho S c \tau) \frac{F^2}{S^2 \rho E} = \frac{F^2 c \tau}{2 S E} = \frac{F^2 \tau}{2 S \sqrt{\rho E}}$

			Potential energy: it equals the kinetic energy in a progressive elastic wave.

			$U = K = \frac{F^2 \tau}{2 S \sqrt{\rho E}}$

			Total energy:

			$\boxed{W' = K + U = \frac{F^2 \tau}{S \sqrt{\rho E}}}$
		@@ -142,7 +142,7 @@Solution

			#### Answer

		−	~~123~~by resuming of the answers have
		+	by resuming of the answers have

			$\boxed{v = \frac{F}{S\sqrt{\rho E}}}$

			$\boxed{p = 0.5 F\tau}$

			$\boxed{W = K + U = \frac{F^2 \tau}{2 S \sqrt{\rho E}}}$

			$\boxed{p' = F\tau}$

			$\boxed{W' = K + U = \frac{F^2 \tau}{S \sqrt{\rho E}}}$
		unchanged lines 7