# 1D Particle & Energy w/ F(x): Am I doing this right?

• monsterhart
In summary: C=\frac{5\cos\left(\frac{\pi}{2}\right)}{2}+C=C\implies C=0.$$In summary, π is not a number. You are not allowed to enter symbols and equations like ##\frac{\pi}{8}##. Use a decimal number for your answers. π is approximately equal to 0.392699081699. monsterhart Homework Statement A particle with a mass of 2.00 kg travels along the positive x axis in the positive x direction under the influence of a periodic conservative force ##F(x) = Fo sin(ax)##, where ##Fo = 10.0\ N## and ##a = 4.00\ m^{−1}##. It starts from the origin with a kinetic energy of 5.00 J and a potential energy of 2.50 J. A) Determine the position ##x_{1}##, where the particle experiences a maximum force for the first time. HINT: The maximum force is when ##sin(ax)=1##, but make sure you are in radians. Enter a decimal number. B) Determine an expression for the potential energy as a function of x. (Use the following as necessary: x. Do not include units in your expression.) C) Determine the speed ##v_{1}## of the particle at the position ##x_{1}##, when it experiences a maximum force for the first time. Enter a decimal number. Relevant Equations$$\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 xTE_{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}+C2.5=\frac{5\cos\left(0\right)}{2}+C\implies 2.5-2.5=C=0U\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\ ma=\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}$$P.S. My attempts can also be found in this calculator sheet: https://www.desmos.com/calculator/ybwhmzhjdj Edit: I submitted my attempt. Does anyone know where I went wrong with Part C? Last edited: Part A This$$\sin\left(4x_{1}\right)=1\implies x_{1}=\frac{\arcsin\left(1\right)}{4}\approx 0.392699081699\ m$$is nonsense. You cannot say that something is approximately equal to a number with 12 decimal places. Furthermore, you will be better off if you realized that ##\arcsin(1)=\frac{\pi}{2}## and used the symbol instead of its numerical value. Part B You say$$U\left(x\right)=-\int F\left(x\right)dx=-10\int\left(\sin\left(4x\right)\right)dx=10\cos\left(4x\right)+C$$That is incorrect.$$\text{Is}~~\frac{d}{dx}\left[10\cos\left(4x\right)+C\right]=-10\sin(4x)~?$$I suggest that you fix that first, then tackle the next part. kuruman said: Part A This$$\sin\left(4x_{1}\right)=1\implies x_{1}=\frac{\arcsin\left(1\right)}{4}\approx 0.392699081699\ m$$is nonsense. You cannot say that something is approximately equal to a number with 12 decimal places. Furthermore, you will be better off if you realized that ##\arcsin(1)=\frac{\pi}{2}## and used the symbol instead of its numerical value. Part B You say$$U\left(x\right)=-\int F\left(x\right)dx=-10\int\left(\sin\left(4x\right)\right)dx=10\cos\left(4x\right)+C$$That is incorrect.$$\text{Is}~~\frac{d}{dx}\left[10\cos\left(4x\right)+C\right]=-10\sin(4x)~?$$I suggest that you fix that first, then tackle the next part. Part A: π is not a number. We are not allowed to enter symbols, nor unsolved equations like ##\frac{\pi}{8}##. Only a decimal number will be accepted. Is there some other reason why it's "nonsense?" Part B: You're right. I was thinking in terms of ##\frac{d}{dx}\left(\cos\left(\theta\right)\right)=-sin\left(\theta\right)## (how it was first taught to me in school) but I suppose it's more like ##\frac{d}{dx}\left(\cos\left(f\left(x\right)\right)\right)=-\sin \left(f\left(x\right)\right)\frac{d}{dx}\left(f\left(x\right)\right)## (how it was taught later after I'd already memorized it the limited way that was originally taught to me). So that'd make it:$$\frac{5\cos\left(4x\right)}{2}+C$$Thanks for the response. I'll update the post for clarity. monsterhart said: Is there some other reason why it's "nonsense?" Yes, you cannot use the "approximately equals" with a distance in meters that is quoted to 12 decimal places. The size of an atom at about is 10-10 m is 100 times bigger. If you must use numbers for your answers, they should be quoted to 3 significant figures which is the accuracy of your input values. A word of advice to make your calculations easier. Note that the force is a maximum when ##\sin(ax)## is maximum and has the value of ##1##. This occurs for the first time when ##ax_1=\frac{\pi}{2}\implies x_1=\frac{\pi}{2a}.## Also note that the integration constant ##C## can be left as is unless you are told specifically where the location of zero of potential energy. For example, if you choose the potential energy to be zero, where the force is maximum at ##x_1##, then$$0=U(x_1)=\frac{5\cos\left(ax_1\right)}{2}+C=\frac{5\cos\left(\frac{\pi}{2}\right)}{2}+C=C\implies C=0.$$Your problem with part C is that you have a nonsense equation. You say$$7.5\ J=v_{f}^{2}+\frac{5\pi}{8}J=v_{f}^{2}+\frac{5\pi}{8}J Does speed squared have units of Joules? The total energy of 7.5 J is the sum of potential and kinetic energy. Start from there and don't forget to adjust the integration constant ##C##, i.e. define the zero of potential energy, so that ##U(0)=2.5~##J.

MatinSAR and berkeman
Ah I see I forgot to write in the kg on the ##v_{f}^{2}##. I updated it. By the way, the input values are given/known values. There are no measured values in this problem, thus no limit to significant digits.

Last edited:
monsterhart said:
There are no measured values in this problem, thus no limit to significant digits.
It doesn't matter if they are measured or not. The given values that you have to work with are to 3 significant figures and that's what counts.
monsterhart said:
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.
Also, please don't edit work that you have already posted. It's confusing for people who try to follow the thread. Fix any problems and post again.

MatinSAR
That's not how significant digits work. When something is measured it is rounded to the most precise digit attainable, making that the significant digit, as digits beyond that could vary. When a number is not a measured value the value is considered to be exact.

A given/known value of "1" means that all digits after the decimal place are definitely 0, so whether it is written 1.0 or 1.00 doesn't matter. Also your logic is inconsistent. There's a "2" in the equation too, so if significant digits depended on the digits of any number used regardless of whether it's a known or measured number the answer would need to be reduced to 1 significant digit, not 3.

I never edited the block you're quoting, nor did I remove anything. I only added details at your prompting for clarity so as to avoid further confusion, and crossed out the mistake in Part B rather than removing it so that it would be clear that the change was made.

I found out the equation it was looking for for Part C (##SPE_{f}## should have been eliminated as well), so I wont be needing further assistance. Thanks again.

## What is the basic concept of a 1D particle in a potential energy field F(x)?

In one-dimensional physics, a particle moving in a potential energy field F(x) can be described by its position and energy as functions of time. The potential energy field F(x) represents the force acting on the particle at any position x, and it influences the particle's motion according to Newton's second law of motion.

## How do I determine the potential energy function F(x) for a given force?

The potential energy function U(x) is related to the force F(x) by the relationship F(x) = -dU(x)/dx. To find U(x), you integrate the force function F(x) with respect to x, and include a constant of integration if necessary. Mathematically, U(x) = -∫F(x) dx + C, where C is a constant.

## What are the common methods for solving the motion of a 1D particle in a potential energy field?

Common methods include solving the differential equations of motion using analytical techniques if the potential energy function is simple, or using numerical methods such as the Euler method, Runge-Kutta methods, or other computational techniques for more complex potentials. Additionally, energy conservation principles can be applied to solve for the particle's velocity and position over time.

## How do I check if my solution to the 1D particle motion problem is correct?

To verify your solution, you can check if it satisfies the initial conditions and the governing differential equations. Additionally, ensure that energy conservation holds, meaning the total mechanical energy (kinetic plus potential) remains constant if no non-conservative forces are present. Comparing your solution against known analytical solutions or numerical simulations can also provide validation.

## What are the typical boundary conditions for a 1D particle in a potential energy field?

Typical boundary conditions include specifying the particle's initial position and velocity at a given time, usually t = 0. In some problems, you might also encounter boundary conditions at the edges of a domain, such as reflective or absorbing boundaries, depending on the physical scenario being modeled.

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