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- Homework Statement
- A particle with a mass of 2.00 kg travels along the positive x axis in the positive x direction under the influence of a periodic conservative force ##F(x) = Fo sin(ax)##, where ##Fo = 10.0\ N## and ##a = 4.00\ m^{−1}##. It starts from the origin with a kinetic energy of 5.00 J and a potential energy of 2.50 J.

A) Determine the position ##x_{1}##, where the particle experiences a maximum force for the first time. HINT: The maximum force is when ##sin(ax)=1##, but make sure you are in radians. Enter a decimal number.

B) Determine an expression for the potential energy as a function of x. (Use the following as necessary: x. Do not include units in your expression.)

C) Determine the speed ##v_{1}## of the particle at the position ##x_{1}##, when it experiences a maximum force for the first time. Enter a decimal number.

- Relevant Equations
- $$\sin\left(4x_{1}\right)=1$$

$$\Delta U(x)=-W_{cons}=-\int_{x_{1}}^{x_{2}}F\left(x\right)dx\approx-F(x)\Delta x$$

$$TE_{i}=\frac{mv_{f}^{2}}{2}+\frac{kx_{1}^{2}}{2}$$

$$k=\frac{ma}{x_{1}}$$

A) I just did what it said to do:

$$\sin\left(4x_{1}\right)=1\implies x_{1}=\frac{\arcsin\left(1\right)}{4}\ m=\frac{\pi}{8}\ m\approx 0.392699081699\ m$$

B) I modified the method from an example from the lecture the other week:

$$U\left(x\right)=-\int F\left(x\right)dx=-10\int\left(\sin\left(4x\right)\right)dx=\rlap{-------}10\cos\left(4x\right)+C$$

$$\rlap{---------------------}2.5=10\cos\left(0\right)+C\implies 2.5-10=C=-7.5$$

$$\rlap{-----------}U\left(x\right)=10\cos\left(4x\right)-7.5$$

$$=\frac{5\cos\left(4x\right)}{2}+C$$

$$2.5=\frac{5\cos\left(0\right)}{2}+C\implies 2.5-2.5=C=0$$

$$U\left(x\right)=\frac{5\cos\left(4x\right)}{2}$$

C) I'm basically just guessing the equation based on the lack of y component:

$$7.5\ J=\frac{2v_{f}^{2}\ kg}{2}+\frac{kx_{1}^{2}}{2}m^{2}=v_{f}^{2}\ kg+\frac{a\pi}{8}kg\ m$$

$$a=\frac{10\sin\left(\frac{4\arcsin\left(1\right)}{4}\right)\ N}{2\ kg}=5\frac{m}{s^{2}}$$

$$7.5\ J=v_{f}^{2}\ kg+\frac{5\pi}{8}J=v_{f}^{2}\ kg+\frac{5\pi}{8}J$$

$$\implies v_{f}=\sqrt{7.5-\frac{5\pi}{8}}\frac{m}{s}=\frac{\sqrt{5}\sqrt{12-\pi}}{2\sqrt{2}}\approx 2.35297781365\frac{m}{s}$$

P.S. My attempts can also be found in this calculator sheet: https://www.desmos.com/calculator/ybwhmzhjdj

Edit: I submitted my attempt. Does anyone know where I went wrong with Part C?

$$\sin\left(4x_{1}\right)=1\implies x_{1}=\frac{\arcsin\left(1\right)}{4}\ m=\frac{\pi}{8}\ m\approx 0.392699081699\ m$$

B) I modified the method from an example from the lecture the other week:

$$U\left(x\right)=-\int F\left(x\right)dx=-10\int\left(\sin\left(4x\right)\right)dx=\rlap{-------}10\cos\left(4x\right)+C$$

$$\rlap{---------------------}2.5=10\cos\left(0\right)+C\implies 2.5-10=C=-7.5$$

$$\rlap{-----------}U\left(x\right)=10\cos\left(4x\right)-7.5$$

$$=\frac{5\cos\left(4x\right)}{2}+C$$

$$2.5=\frac{5\cos\left(0\right)}{2}+C\implies 2.5-2.5=C=0$$

$$U\left(x\right)=\frac{5\cos\left(4x\right)}{2}$$

C) I'm basically just guessing the equation based on the lack of y component:

$$7.5\ J=\frac{2v_{f}^{2}\ kg}{2}+\frac{kx_{1}^{2}}{2}m^{2}=v_{f}^{2}\ kg+\frac{a\pi}{8}kg\ m$$

$$a=\frac{10\sin\left(\frac{4\arcsin\left(1\right)}{4}\right)\ N}{2\ kg}=5\frac{m}{s^{2}}$$

$$7.5\ J=v_{f}^{2}\ kg+\frac{5\pi}{8}J=v_{f}^{2}\ kg+\frac{5\pi}{8}J$$

$$\implies v_{f}=\sqrt{7.5-\frac{5\pi}{8}}\frac{m}{s}=\frac{\sqrt{5}\sqrt{12-\pi}}{2\sqrt{2}}\approx 2.35297781365\frac{m}{s}$$

P.S. My attempts can also be found in this calculator sheet: https://www.desmos.com/calculator/ybwhmzhjdj

Edit: I submitted my attempt. Does anyone know where I went wrong with Part C?

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