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+### Statement
+
+$3.7.6.$ [Insert the problem statement]
+
+### Solution
+
+Longitudinal waves in a thin bar propagate with speed
+
+\boxed{c = \sqrt{\frac{E}{\rho}}}
+
+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. 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}}}
+
+
+#### Answer
+
+[Insert a concise answer or boxed result]