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en/2.3.42.md
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| + | ### Statement | ||
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| + | $2.3.42.$ [Insert the problem statement] | ||
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| + | ### Solution | ||
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| + | 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 $v_{\mbox{max}}=\omega A=g\sqrt{m/k}$. | ||
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| + | #### Answer | ||
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| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $2.3.42.$ [Insert the problem statement] | |||
| ### 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] | |||