Answer :
Answer:
[tex]\begin{gathered} v_{\max }=\sqrt[]{25\times k\Delta x^2} \\ \Delta x^2=\text{ Max Stretch} \end{gathered}[/tex]Explanation:
We need to find the max and min velocity conditions. In the spring-mass system harmonic oscillator, we have the following conditions.
(1) Maximum stretched
At a maximum stretch, kinetic energy is 0 therefore 0 velocity and potential energy is maximum
[tex]\begin{gathered} KE=\frac{1}{2}mv^2=0 \\ PE=\frac{1}{2}k\Delta x^2=\text{Max} \\ \therefore\Rightarrow \\ ME=KE+PE\Rightarrow\frac{1}{2}mv^2+\frac{1}{2}k\Delta x^2 \\ ME=0+\frac{1}{2}k\Delta x^2 \\ ME=\frac{1}{2}k\Delta x^2 \\ \end{gathered}[/tex]Note! that maximum stretch can be negative and positive, depending on the direction.
(2) Equlibrium position/ no stretch
[tex]\begin{gathered} KE=\frac{1}{2}mv^2=\text{Max} \\ PE=\frac{1}{2}k\Delta x^2=0 \\ \therefore\Rightarrow \\ M.E=KE+PE\Rightarrow\frac{1}{2}mv^2+\frac{1}{2}k\Delta x^2 \\ M.E=\frac{1}{2}mv^2+0 \\ \text{Therefore, the Max/Min velocity can be found using these two conditions (1) and (2)} \\ v_{\max }=\sqrt[]{\frac{2\times M.E}{m}} \\ v_{\max }=\sqrt[]{\frac{2\times M.E}{100}}=\sqrt[]{50\times M.E} \\ \text{SInce Mechanical energy is constant, therefore} \\ v_{\max }=\sqrt[]{25\times k\Delta x^2} \\ \text{Where: } \\ \Delta x=\text{ Max Stretch} \\ \\ \\ \end{gathered}[/tex]