Правка разделов «Statement», «Answer»
en/2.3.42.md
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| ### Statement | |||
| − | $2.3.42.$ [Insert the problem statement] | ||
| + | $2.3.42.$ A weight of mass $m$ suspended on a spring of stiffness $k$ is placed on a stand. The spring is not deformed. The stand is quickly removed. Determine the maximum spring extension and maximum load speed. | ||
| ### Solution | |||
| After the stand is removed, the object will undergo simple harmonic motion with angular frequency $\omega=\sqrt{k/m}$ with its initial position as the highest position, because the object must have zero speed at this point. The equilibrium point will be at the distance $A$ below the initial position so that $kA=mg$, i.e. $A=mg/k$. This is the amplitude of the motion, and so the maximum elongation of the spring is $x_{\mbox{max}}=2A=2mg/k$. The maximum speed $v_{\mbox{max}}$ is attained at the equilibrium point, and $v_{\mbox{max}}=\omega A=g\sqrt{m/k}$. | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $x_{\mbox{max}}=2mg/k$ | ||
| + | |||
| + | $v_{\mbox{max}}=g\sqrt{m/k}$ | ||
| @@ -1,11 +1,13 @@ | |||
| ### Statement | ### Statement | ||
| $2.3.42.$ [Insert the problem statement] | $2.3.42.$ A weight of mass $m$ suspended on a spring of stiffness $k$ is placed on a stand. The spring is not deformed. The stand is quickly removed. Determine the maximum spring extension and maximum load speed. | ||
| ### Solution | ### Solution | ||
| After the stand is removed, the object will undergo simple harmonic motion with angular frequency $\omega=\sqrt{k/m}$ with its initial position as the highest position, because the object must have zero speed at this point. The equilibrium point will be at the distance $A$ below the initial position so that $kA=mg$, i.e. $A=mg/k$. This is the amplitude of the motion, and so the maximum elongation of the spring is $x_{\mbox{max}}=2A=2mg/k$. The maximum speed $v_{\mbox{max}}$ is attained at the equilibrium point, and $v_{\mbox{max}}=\omega A=g\sqrt{m/k}$. | After the stand is removed, the object will undergo simple harmonic motion with angular frequency $\omega=\sqrt{k/m}$ with its initial position as the highest position, because the object must have zero speed at this point. The equilibrium point will be at the distance $A$ below the initial position so that $kA=mg$, i.e. $A=mg/k$. This is the amplitude of the motion, and so the maximum elongation of the spring is $x_{\mbox{max}}=2A=2mg/k$. The maximum speed $v_{\mbox{max}}$ is attained at the equilibrium point, and $v_{\mbox{max}}=\omega A=g\sqrt{m/k}$. | ||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $x_{\mbox{max}}=2mg/k$ | ||
| $v_{\mbox{max}}=g\sqrt{m/k}$ | |||