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In some diatomic molecules, the force each atom exerts on the other can be approximated by $F=-C / r^{2}+D / r^{3}$ , where $r$ is the atomic separation and $C$ and $D$ are positive constants. $(a)$ Graph $F$ vs. $r$ from $r=0.8 D / C$ to $r=4 D / C$ . (b) Show that equilibrium occurs at $r=r_{0}=D / C$ . (c) Let $\Delta r=r-r_{0}$ be a small displacement from equilibrium, where $\Delta r \ll r_{0} .$ Show that for such small displacements, the motion is approximately simple harmonic, and (d) deter- mine the force constant. (e) What is the period of such motion? [Hint: Assume one atom is kept at rest.]

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a. graph not availableb. $$r_{0}=\frac{D}{C}$$c. $$F=\frac{-C \Delta r}{\left(r_{0}+\Delta r\right)^{3}}-\left(\frac{C}{r_{0}^{3}}\right) \Delta r$$d. $$\frac{C^{4}}{D^{3}}$$e. $$T=2 \pi\left(\frac{m D^{3}}{C^{4}}\right)^{\frac{1}{2}}$$

Physics 101 Mechanics

Chapter 14

Oscillators

Motion Along a Straight Line

Motion in 2d or 3d

Periodic Motion

Rutgers, The State University of New Jersey

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Lectures

04:01

2D kinematics is the study of the movement of an object in two dimensions, usually in a Cartesian coordinate system. The study of the movement of an object in only one dimension is called 1D kinematics. The study of the movement of an object in three dimensions is called 3D kinematics.

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support. They asked us the graph. The force is a function of our Ah, and this would be the graph here. You can use Excel or Matt Lab or any, uh, graphing software in order Thio model this and then for part B. Let's figure out the equilibrium position. So for part B Ah, we can say that the force are not equals negative, See, over are not squared. Plus de over are not cute at this point. We know that this is gonna equal zero because we want to find the equilibrium position. Therefore see over our squared equals d over R cubed And so essentially the equilibrium position are not This will equal d over. See, So this would be our answer for part B. Now, for part C, we're trying to find the net force, uh, for party we're trying to find in that force at, uh are equaling are not the equilibrium position. Plus, the change in our delts are so we consider it af of r plus the delta are would be equal to negative C times are not plus doubts are gone. We're simply just ah substituting. And so at this point, we can say that half of our plus change and are would be equal to negative. See, over are not squared. Times one minus two Delta are divided by our not so we have to make an approximation plus de over r cubed. And then this would be one minus three delta are over are not. And so, uh, at this point, we can, uh, further simplify and say that this is gonna be equal to see over are not cubed plus negative are not. Times one minus to delta are over are not plus de over sea And then this will be again one minus three dots are over All right, We can say that half of our plus delta r equals C over are not cute And then this will be minus on our notch plus two times out to our plus are not minus three times Delta are And we can cancel out these and say that f r plus Delta are will equal negative psi Delta are over r cubed and it's important to note that here the force is going to be directly proportional to the change in position. Um, however, it's gonna be in the opposite direction. So force is directly proportional to the change in position and in the opposite direction. Uh, this would again, uh, model simple harmonic motion or a simple harmonic oscillator. So at this point, for simple harmonic motion for part D four simple harmonic motion, we know that force equals negative K X, Therefore, K is simply equal to see over aren't not cubed. And this leak will see to the fourth over D Cube, given that are not equals d oversee. So we plug this into here and we Cubitt, and this would be equal to the spring constant. So this would be your answer here and then for part E Lastly, we want to find the period. So for E, the period is gonna equal to pi times the square root of the mass divided by the spring spring constant. So this would be equal to two pi times. The mass times d cubed, divided by see to the fourth. And so this would equal the t the period. So it would be our final answer for period for party. That is the end of the solution. Thank you for watching

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