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 phy131studiof15:lectures:chapter16 [2015/07/22 11:00]mdawber created phy131studiof15:lectures:chapter16 [2015/10/31 12:15] (current)mdawber [Energy in SHM] 2015/10/31 12:15 mdawber [Energy in SHM] 2015/10/31 12:12 mdawber [Simple Pendulum] 2015/10/31 12:09 mdawber [Equations of motion for SHM] 2015/07/22 11:03 mdawber [A vertical spring] 2015/07/22 11:03 mdawber [A vertical spring] 2015/07/22 11:02 mdawber [Simple harmonic motion with multiple springs] 2015/07/22 11:00 mdawber created Next revision Previous revision 2015/10/31 12:15 mdawber [Energy in SHM] 2015/10/31 12:12 mdawber [Simple Pendulum] 2015/10/31 12:09 mdawber [Equations of motion for SHM] 2015/07/22 11:03 mdawber [A vertical spring] 2015/07/22 11:03 mdawber [A vertical spring] 2015/07/22 11:02 mdawber [Simple harmonic motion with multiple springs] 2015/07/22 11:00 mdawber created Line 13: Line 13: {{shmvspring.png}} {{shmvspring.png}} + When the system is in equilibrium + + $kx_{0}=mg$ + + so + + $x_{0}=\frac{mg}{k}$ ===== SHM as a function of time ===== ===== SHM as a function of time ===== Line 42: Line 49: $T=2\pi\sqrt{\frac{m}{k}}$ ​ $T=2\pi\sqrt{\frac{m}{k}}$ ​ + + ===== 16.P.003 ===== + + ===== 16.P.008 ===== + + ===== 16.P.021 ===== + + + ===== Energy in SHM ===== ===== Energy in SHM ===== Line 50: Line 66: $v=\pm\sqrt{\frac{k}{m}(A^{2}-x^{2})}$ $v=\pm\sqrt{\frac{k}{m}(A^{2}-x^{2})}$ + + ===== 16.P.044 ===== ===== Simple harmonic motion with multiple springs ===== ===== Simple harmonic motion with multiple springs ===== Line 65: Line 83: + + ===== Energy for the vertical spring ===== + + A vertical spring will also execute simple harmonic motion, though it's mean position will be modified by the balance of the gravitational force and the spring force. + + {{shmvspring.png}} + + When the system is in equilibrium + + $kx_{0}=mg$ + + so + + $x_{0}=\frac{mg}{k}$ + + + There are two sources of potential energy in the vertical spring, the energy stored in the spring and gravitational potential energy. We can define the zero of total potential energy, U, as the equilibrium position + + $U=\frac{1}{2}k(x-x_{0})^{2}+mgx-\frac{1}{2}kx_{0}^{2}$ + + $U=\frac{1}{2}kx^{2}-kxx_{0}+mgx$ + + Using our previous result for $x_{0}$ + + $U=\frac{1}{2}kx^{2}-mgx+mgx$ + + $U=\frac{1}{2}kx^{2}$ Line 87: Line 132: [[https://​www.youtube.com/​watch?​v=kBqjbZeAN9g|Video of many carefully arranged pendula]] [[https://​www.youtube.com/​watch?​v=kBqjbZeAN9g|Video of many carefully arranged pendula]] + + ===== 16.P.029 =====