Consider a spring fixed at its upper end and supporting a weight of 10 pounds at its lower end. Suppose the 10-pound weight stretches the spring by 6 inches. Find the equation of motion of the weight if it is drawn to a position 4 inches below its equilibrium position and released.

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Accepted Answer

By Hooke's law, since a force of [latex]10 \mathrm{lb}[/latex] stretches the spring by [latex]\frac{1}{2} \mathrm{ft}[/latex], [latex]10=k\left(\frac{1}{2}\right)[/latex] or [latex]k=20 \mathrm{lb} / \mathrm{ft}[/latex]. We are given the initial values [latex]x_{0}=\frac{1}{3} \mathrm{ft}[/latex] and [latex]v_{0}=0[/latex], so by equation (3) and the identity [latex]^{\ddagger} k / m=g k / w=64 \sec ^{-2}[/latex], we obtain

[latex]x(t)=x_{0} \cos \omega_{0} t+\left(v_{0} / \omega_{0}\right) \sin \omega_{0} t[/latex]

[latex]x(t)=\frac{1}{3} \cos 8 t \mathrm{ft}[/latex].
Thus the amplitude is [latex]\frac{1}{3} \mathrm{ft}(=4[/latex] in. [latex])[/latex], and the frequency is [latex]f=4 / \pi[/latex] hertz.

Question 11.13.3: Consider a spring fixed at its upper end and supporting a we...

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