Vertical Spring–Mass System
Natural Length, Mean Position and Extreme Position of a Vertical Spring
Any Position
The spring obeys Hooke’s Law:
\[F_{\text{spring}} = -kx\] \[ma = -kx\] \[a = -\frac{k}{m}x\] \[a = -\omega^2 x\] \[\boxed{\omega = \sqrt{\frac{k}{m}}\frac{k}{m}}\]Mean Position
At equilibrium:
\[F_{\text{spring}} = F_{\text{gravity}}\] \[kx_{\text{mean}} = mg\] \[\boxed{x_{\text{mean}}=\frac{mg}{k}}\]Extreme Position
Using conservation of energy:
\[W_{\text{gravity}}=PE_{\text{spring}} +KE\] \[mgx_{\text{extreme}}=\frac{1}{2}kx_{\text{extreme}}^2 + 0\] \[\boxed{x_{\text{extreme}}=\frac{2mg}{k}}\]Vertical Spring SHM with Shifted Equilibrium
Simple Harmonic Motion of a Vertical Spring–Mass System
Forces on the Mass
Net restoring force:
\[F_{\text{net}}=-\left(F_{\text{spring}}-F_{\text{gravity}}\right)\] \[F_{\text{net}}=-(kx - mg)\]At equilibrium:
\[mg = kx_{\text{mean}}\] \[F_{\text{net}}=-(kx - kx_{\text{mean}})\] \[ma = -k(x - x_{\text{mean}})\]Using shifted coordinate:
\[X = x - x_{\text{mean}}\]Equation of SHM
\[a = -\frac{k}{m}X\] \[a = -\omega^2 X\] \[\boxed{\omega=\sqrt{\dfrac{k}{m}}\dfrac{k}{m}}\]Velocity at Mean Position
Using conservation of energy:
\[PE_{\text{spring}} + KE_{\text{mean}} = U_i - U_f\]
\[\frac{1}{2}kx_{\text{mean}}^{2} + \frac{1}{2}mv_{\text{mean}}^{2} = mgx_{\text{mean}}\]
\[\frac{1}{2}k\left(\frac{mg}{k}\right)^2 + \frac{1}{2}mv_{\text{mean}}^{2}= mg\left(\frac{mg}{k}\right)\]
\[\frac{1}{2}mv_{\text{mean}}^{2}= \frac{1}{2}\frac{m^{2}g^{2}}{k}\]
\[v_{\text{mean}} = g\sqrt{\frac{m}{k}}\frac{m}{k}\]
\[\boxed{v_{\max}=g\sqrt{\frac{m}{k}}\frac{m}{k}}\]
No comments:
Post a Comment
Please provide your valuable feedback. Students, Parents, Teachers.