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Solution:- given oscillating springs differential equations.(i) 9 s^{\prime \prime}+s=0, \quad s(0)=10, s^{\prime}(0)=0(ii) 16 S^{\prime \prime}+S=0, s(0)=20, s^{\prime}(0)=0(iii) S^{\prime \prime}+9 S=0, \quad S(0)=5, S^{\prime}(0)=0(iv) s^{\prime \prime}+16 s=0, \quad s(0)=4, \quad s^{\prime}(0)=0we know that general oscillating springs equation\frac{d^{2} s}{d t^{2}}+w^{2} s=0Where Angulor velocity =wAmplitude A=S(0)maximum velocity V_{m_{x}}=\omega ATime period (T)=\frac{2 \pi}{\omega}for given equations.(i)S^{\prime \prime}+\frac{1}{9} S=0, S(0)=10, S^{\prime}(0)=0\omega^{2}=\frac{1}{9} \quad \omega=\frac{1}{3}Amplitude A=S(0)=10\begin{array}{l}V_{\text {max }}=\omega A=\frac{10}{3} \mathrm{~m} / \mathrm{s} \\T=\frac{2 \pi}{\omega}=\frac{2 \pi}{1 / 3}=6 \pi \mathrm{sec}\end{array}(ii)\begin{array}{l}S^{\prime \prime}+\frac{1}{16} S=0, S(0)=20, \quad S^{\prime}(0)=0 \\\omega^{2}=\frac{1}{16} \quad \omega=\frac{1}{4} \\A=S(0)=20 \quad A=20 \\V_{\max }=\omega A=\frac{20}{4}=5 \mathrm{~m} / \mathrm{s} \\T=\frac{2 \pi}{\omega}=\frac{2 \pi}{1 / 4}=8 \pi \mathrm{sec}\end{array}(iii)\begin{array}{l}S^{\prime \prime}+9 S=0, S(0)=5, \quad S^{\prime}(0)=0 \\\omega^{2}=9 \quad \omega=3 \\A=S(0) \quad A=5 \\V_{\text {max }}=\omega A \quad V_{\text {max }}=15 \mathrm{~m} / \mathrm{sec}\end{array}T=\frac{2 \pi}{\omega} \quad T=\frac{2 \pi}{3} \mathrm{sec}(iv)\begin{array}{l}S^{\prime \prime}+16 S=0, S(0)=4, \quad S^{\prime}(0)=0 \\\omega^{2}=16 \quad \omega=4 \\A=S(0) \quad A=4 \\V_{\text {max }}=\omega A \quad V_{\text {max }}=16 \mathrm{~m} / \mathrm{sec} \\T=\frac{2 \pi}{\omega} T=\frac{2 \pi}{4}=\frac{\pi}{2} 5 \mathrm{cc}\end{array}By comparing above results.a. T is minimums for=(iv).So (iv) spring oscillating most quickly.b. The spring oscillating with largest Amplitude = (ii)(C) The spring oscillating most slowly = (ii)(d) The spring oscillating with the largest maximum velocites = ivAns. ...