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| ### Statement |
| ### Statement |
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| $3.7.6.$ [Insert the problem statement] |
| $3.7.6.$ . |
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| A longitudinal force F acts on the end of a resting semi-infinite rod for time |
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| τ. Find the velocity of the rod particles and its deformation in the area of the |
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| resulting wave if the cross-section of the rod is S, the Young’s modulus of its |
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| material is E, and the density is ρ. What is the density of the rod in the wave |
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| region? Find the momentum and energy of the displaced particles of the rod |
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| after time 0.5τ and 1.5τ from the beginning of the force. |
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| ### Solution |
| ### Solution |
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| Longitudinal waves in a thin bar propagate with speed |
| The longitudinal waves in a thin bar propagate with speed |
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| \boxed{c = \sqrt{\frac{E}{\rho}}} |
| $\boxed{c = \sqrt{\frac{E}{\rho}}}$ |
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| This speed does not depend on the sign of the force. |
| and this speed does not depend on the sign of the force. |
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| The compressive force produces a negative normal stress (compression): |
| The compressive force produces a negative normal stress (compression): |
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| \sigma = -\frac{F}{S} |
| $\sigma = -\frac{F}{S}$ |
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| By Hooke's law, the unit strain is \varepsilon = \sigma / E: |
| By Hooke's law, the unit strain is $\varepsilon = \sigma / E$: |
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| \boxed{\varepsilon = -\frac{F}{SE}} |
| $\boxed{\varepsilon = -\frac{F}{SE}}$ |
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| The quantity |\varepsilon| is the relative shortening. The negative sign indicates compression. |
| The quantity $|\varepsilon|$ is the relative shortening. |
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| and the negative sign indicates compression. |
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| In a progressive plane elastic wave traveling in the positive direction, the relationship between the stress \sigma and the particle velocity v is: |
| In a progressive plane elastic wave traveling in the positive direction, the relationship between the stress $\sigma $ and the particle velocity v is: |
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| \sigma = -\rho c v |
| $\sigma = -\rho c v$ |
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| (the minus sign appears because in compression the material moves in the direction of wave propagation). Solving for v: |
| (the minus sign appears because in compression the material moves in the direction of wave propagation). Solving for v: |
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| v = -\frac{\sigma}{\rho c} = -\frac{(-F/S)}{\rho c} = \frac{F}{S\rho c} |
| $v = -\frac{\sigma}{\rho c} = -\frac{(-F/S)}{\rho c} = \frac{F}{S\rho c}$ |
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| Substituting c = \sqrt{E/\rho}: |
| Substituting $c = \sqrt{E/\rho}$: |
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| \boxed{v = \frac{F}{S\sqrt{\rho E}}} |
| $\boxed{v = \frac{F}{S\sqrt{\rho E}}}$ |
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| The particles move in the direction of wave propagation, with constant speed as long as the force is applied. |
| The particles move in the direction of wave propagation, with constant speed as long as the force is applied. |
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| 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: |
| Conservation of mass implies that the density \rho' in the deformed region satisfies $\rho' = \rho / (1+\varepsilon)$. |
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| \rho' \approx \rho(1 - \varepsilon) = \rho\left(1 + \frac{F}{SE}\right) |
| In the linear regime $(|\varepsilon| \ll 1)$ we can approximate: |
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| Since \varepsilon is negative, the density increases in the compressed region. |
| $\rho' \approx \rho(1 - \varepsilon) = \rho\left(1 + \frac{F}{SE}\right)$ |
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| \boxed{\rho' = \rho\left[1 + \frac{F}{SE}\right]} |
| Since $\varepsilon $ is negative, the density increases in the compressed region. |
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| 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: |
| $\boxed{\rho' = \rho\left[1 + \frac{F}{SE}\right]}$ |
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| m(t) = \rho S l(t) = \rho S c t |
| While the force is applied$ (0 < t \le \tau)$, |
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| the wave front advances a distance $l(t) = ct$. |
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| The total mass set into motion up to that instant is: |
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| Momentum and energy at t = 0.5\tau |
| $m(t) = \rho S l(t) = \rho S c t% |
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| At this instant the force is still acting; the disturbance has not yet ceased. |
| Momentum and energy at$ t = 0.5\tau$ |
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| · Mass in motion: m = \rho S c \cdot 0.5\tau. |
| At this instant the force is still acting; |
| · Impulse: p = m v. |
| the disturbance has not yet ceased. |
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| 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}} |
| Mass in motion: $m = \rho S c \cdot 0.5\tau.$ |
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| Impulse:$ p = m v$. |
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| Since \rho c = \rho\sqrt{E/\rho} = \sqrt{\rho E}, this simplifies to |
| $p = (\rho S c \cdot 0.5\tau) \left( \frac{F}{S\sqrt{\rho E}} \right) = 0.5 F\tau \frac{\rho c}{\sqrt{\rho E}}$ |
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| \boxed{p = 0.5 F\tau} |
| Since |
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| $\rho c = \rho\sqrt{E/\rho} = \sqrt{\rho E}$, |
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| · Kinetic energy: K = \frac{1}{2} m v^2. |
| this simplifies to |
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| 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} |
| $\boxed{p = 0.5 F\tau}$ |
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| Using c/E = 1/\sqrt{\rho E}: |
| Kinetic energy: |
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| $K = \frac{1}{2} m v^2$. |
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| K = \frac{F^2 \tau}{4 S \sqrt{\rho E}} |
| $K = \frac{1}{2} (0.5 \rho S c \tau) \left( \frac{F}{S\sqrt{\rho E}} \right)^2 = \frac{F^2 c \tau}{4 S E}$ |
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| · Elastic potential energy: energy density u_e = \frac{1}{2} \sigma \varepsilon = \frac{F^2}{2 E S^2}. |
| Using $c/E = 1/\sqrt{\rho E}$: |
| Perturbed volume V = S l = 0.5 S c \tau. |
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| 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}} |
| $K = \frac{F^2 \tau}{4 S \sqrt{\rho E}}$ |
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| · Total energy: |
| $Elastic potential energy: energy density u_e = \frac{1}{2} \sigma \varepsilon = \frac{F^2}{2 E S^2}$. |
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| Perturbed volume |
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| $V = S l = 0.5 S c \tau$. |
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| \boxed{W = K + U = \frac{F^2 \tau}{2 S \sqrt{\rho E}}} |
| $U = u_e V = \frac{F^2}{2 E S^2} \cdot 0.5 S c \tau = \frac{F^2 c \tau}{4 E S} = \frac{F^2 \tau}{4 S \sqrt{\rho E}}$ |
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| Momentum and energy at t = 1.5\tau |
| Total energy: |
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| 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: |
| $\boxed{W = K + U = \frac{F^2 \tau}{2 S \sqrt{\rho E}}}$ |
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| m = \rho S L = \rho S c \tau |
| Momentum and energy at $t = 1.5\tau$ |
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| · Impulse: p' = m v. |
| 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: |
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| p' = (\rho S c \tau) \left( \frac{F}{S\sqrt{\rho E}} \right) = F\tau \frac{\rho c}{\sqrt{\rho E}} = F\tau |
| $m = \rho S L = \rho S c \tau$ |
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| \boxed{p' = F\tau} |
| Impulse:$ p' = m v$. |
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| · Kinetic energy: |
| $ p' = (\rho S c \tau) \left( \frac{F}{S\sqrt{\rho E}} \right) = F\tau \frac{\rho c}{\sqrt{\rho E}} = F\tau$ |
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| 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}} |
| $\boxed{p' = F\tau}$ |
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| · Potential energy: it equals the kinetic energy in a progressive elastic wave. |
| Kinetic energy: |
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| U = K = \frac{F^2 \tau}{2 S \sqrt{\rho E}} |
| $K = \frac{1}{2} m v^2 = \frac{1}{2} (\rho S c \tau) \frac{F^2}{S^2 \rho E} = \frac{F^2 c \tau}{2 S E} = \frac{F^2 \tau}{2 S \sqrt{\rho E}}$ |
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| · Total energy: |
| Potential energy: it equals the kinetic energy in a progressive elastic wave. |
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| \boxed{W' = K + U = \frac{F^2 \tau}{S \sqrt{\rho E}}} |
| $U = K = \frac{F^2 \tau}{2 S \sqrt{\rho E}}$ |
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| Total energy: |
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| $\boxed{W' = K + U = \frac{F^2 \tau}{S \sqrt{\rho E}}}$ |
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| #### Answer |
| #### Answer |
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| [Insert a concise answer or boxed result] |
| 123by resuming of the answers have |
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| $\boxed{v = \frac{F}{S\sqrt{\rho E}}}$ |
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| $\boxed{p = 0.5 F\tau}$ |
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| $\boxed{W = K + U = \frac{F^2 \tau}{2 S \sqrt{\rho E}}}$ |
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| $\boxed{p' = F\tau}$ |
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| $\boxed{W' = K + U = \frac{F^2 \tau}{S \sqrt{\rho E}}}$ |
