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| ### Statement |
| ### Statement |
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| $3.7.6.$ . |
| $3.7.6.$ |
| A longitudinal force F acts on the end of a resting semi-infinite rod for time |
| 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 |
| 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 |
| 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 |
| 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 |
| 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. |
| after time 0.5τ and 1.5τ from the beginning of the force. |
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| ### Solution |
| ### Solution |
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| The 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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| and 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 $\varepsilon = \sigma / E$: |
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| By Hooke's law the unit strain is |
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| $\varepsilon = \sigma / E$: |
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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. |
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| The quantity $|\varepsilon|$ |
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| is the relative shortening. |
| and the negative sign indicates compression. |
| 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: |
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| In a progressive plane elastic wave traveling in the positive direction, |
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| 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 $\rho' = \rho / (1+\varepsilon)$. | | Conservation of mass implies that the density \rho' in the deformed region satisfies $\rho' = \rho / (1+\varepsilon)$. |
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| In the linear regime $(|\varepsilon| \ll 1)$ we can approximate: | | In the linear regime $(|\varepsilon| \ll 1)$ we can approximate: |
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| $\rho' \approx \rho(1 - \varepsilon) = \rho\left(1 + \frac{F}{SE}\right)$ | | $\rho' \approx \rho(1 - \varepsilon) = \rho\left(1 + \frac{F}{SE}\right)$ |
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| Since $\varepsilon $ is negative, the density increases in the compressed region. | | Since $\varepsilon $ is negative, the density increases in the compressed region. |
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| $\boxed{\rho' = \rho\left[1 + \frac{F}{SE}\right]}$ | | $\boxed{\rho' = \rho\left[1 + \frac{F}{SE}\right]}$ |
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| While the force is applied$ (0 < t \le \tau)$, | | While the force is applied$ (0 < t \le \tau)$, |
| the wave front advances a distance $l(t) = ct$. | | the wave front advances a distance $l(t) = ct$. |
| The total mass set into motion up to that instant is: | | The total mass set into motion up to that instant is: |
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| $m(t) = \rho S l(t) = \rho S c t% | | $m(t) = \rho S l(t) = \rho S c t% |
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| Momentum and energy at$ t = 0.5\tau$ | | Momentum and energy at$ t = 0.5\tau$ |
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| At this instant the force is still acting; | | At this instant the force is still acting; |
| the disturbance has not yet ceased. | | the disturbance has not yet ceased. |
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| Mass in motion: $m = \rho S c \cdot 0.5\tau.$ | | Mass in motion: $m = \rho S c \cdot 0.5\tau.$ |
| Impulse:$ p = m v$. | | Impulse:$ p = m v$. |
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| $p = (\rho S c \cdot 0.5\tau) \left( \frac{F}{S\sqrt{\rho E}} \right) = 0.5 F\tau \frac{\rho c}{\sqrt{\rho E}}$ | | $p = (\rho S c \cdot 0.5\tau) \left( \frac{F}{S\sqrt{\rho E}} \right) = 0.5 F\tau \frac{\rho c}{\sqrt{\rho E}}$ |
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| Since | | Since |
| $\rho c = \rho\sqrt{E/\rho} = \sqrt{\rho E}$, | | $\rho c = \rho\sqrt{E/\rho} = \sqrt{\rho E}$, |
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| this simplifies to | | this simplifies to |
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| $\boxed{p = 0.5 F\tau}$ | | $\boxed{p = 0.5 F\tau}$ |
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| Kinetic energy: | | Kinetic energy: |
| $K = \frac{1}{2} m v^2$. | | $K = \frac{1}{2} m v^2$. |
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| $K = \frac{1}{2} (0.5 \rho S c \tau) \left( \frac{F}{S\sqrt{\rho E}} \right)^2 = \frac{F^2 c \tau}{4 S E}$ | | $K = \frac{1}{2} (0.5 \rho S c \tau) \left( \frac{F}{S\sqrt{\rho E}} \right)^2 = \frac{F^2 c \tau}{4 S E}$ |
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| Using $c/E = 1/\sqrt{\rho E}$: | | Using $c/E = 1/\sqrt{\rho E}$: |
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| $K = \frac{F^2 \tau}{4 S \sqrt{\rho E}}$ | | $K = \frac{F^2 \tau}{4 S \sqrt{\rho E}}$ |
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| $Elastic potential energy: energy density u_e = \frac{1}{2} \sigma \varepsilon = \frac{F^2}{2 E S^2}$. | | $Elastic potential energy: energy density u_e = \frac{1}{2} \sigma \varepsilon = \frac{F^2}{2 E S^2}$. |
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| Perturbed volume | | Perturbed volume |
| $V = S l = 0.5 S c \tau$. | | $V = S l = 0.5 S c \tau$. |
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| $U = u_e V = \frac{F^2}{2 E S^2} \cdot 0.5 S c \tau = \frac{F^2 c \tau}{4 E S} = \frac{F^2 \tau}{4 S \sqrt{\rho E}}$ | | $U = u_e V = \frac{F^2}{2 E S^2} \cdot 0.5 S c \tau = \frac{F^2 c \tau}{4 E S} = \frac{F^2 \tau}{4 S \sqrt{\rho E}}$ |
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| Total energy: | | Total energy: |
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| $\boxed{W = K + U = \frac{F^2 \tau}{2 S \sqrt{\rho E}}}$ | | $\boxed{W = K + U = \frac{F^2 \tau}{2 S \sqrt{\rho E}}}$ |
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| Momentum and energy at $t = 1.5\tau$ | | Momentum and energy at $t = 1.5\tau$ |
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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: | | 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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| $m = \rho S L = \rho S c \tau$ | | $m = \rho S L = \rho S c \tau$ |
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| Impulse:$ p' = m v$. | | Impulse:$ p' = m v$. |
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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$ | | $ 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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| $\boxed{p' = F\tau}$ | | $\boxed{p' = F\tau}$ |
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| Kinetic energy: | | Kinetic energy: |
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| $K = \frac{1}{2} m v^2 = \frac{1}{2} (\rho S c \tau) \frac{F^2}{S^2 \rho E} = \frac{F^2 c \tau}{2 S E} = \frac{F^2 \tau}{2 S \sqrt{\rho E}}$ | | $K = \frac{1}{2} m v^2 = \frac{1}{2} (\rho S c \tau) \frac{F^2}{S^2 \rho E} = \frac{F^2 c \tau}{2 S E} = \frac{F^2 \tau}{2 S \sqrt{\rho E}}$ |
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| Potential energy: it equals the kinetic energy in a progressive elastic wave. | | Potential energy: it equals the kinetic energy in a progressive elastic wave. |
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| $U = K = \frac{F^2 \tau}{2 S \sqrt{\rho E}}$ | | $U = K = \frac{F^2 \tau}{2 S \sqrt{\rho E}}$ |
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| Total energy: | | Total energy: |
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| $\boxed{W' = K + U = \frac{F^2 \tau}{S \sqrt{\rho E}}}$ | | $\boxed{W' = K + U = \frac{F^2 \tau}{S \sqrt{\rho E}}}$ |
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| #### Answer | | #### Answer |
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| 123by resuming of the answers have | | 123by resuming of the answers have |
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| $\boxed{v = \frac{F}{S\sqrt{\rho E}}}$ | | $\boxed{v = \frac{F}{S\sqrt{\rho E}}}$ |
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| $\boxed{p = 0.5 F\tau}$ | | $\boxed{p = 0.5 F\tau}$ |
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| $\boxed{W = K + U = \frac{F^2 \tau}{2 S \sqrt{\rho E}}}$ | | $\boxed{W = K + U = \frac{F^2 \tau}{2 S \sqrt{\rho E}}}$ |
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| $\boxed{p' = F\tau}$ | | $\boxed{p' = F\tau}$ |
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| $\boxed{W' = K + U = \frac{F^2 \tau}{S \sqrt{\rho E}}}$ | | $\boxed{W' = K + U = \frac{F^2 \tau}{S \sqrt{\rho E}}}$ |
