Applying a time-dependent force to a proton

In summary: Right. Note that this does not return the particle to the origin, but only brings its motion to rest. Essentially, at the beginning of each "cycle" of the force, there should be a constant of integration added representing the offset at the end of the previous cycle.A way to take this into account without breaking your brain is to turn your integrals into definite integrals, hence:$$v(t) = \int_0^t \frac{F_o}{m_p} sin(\omega \tau)\;d\tau$$Then do the same with that result to find the displacement:$$x(t) = \int_0^t \frac{F_o}{m_p \omega} cos(\
  • #1
vbrasic
73
3

Homework Statement



A proton is initially located at the origin of some coordinate system (at rest), when a time-dependent force, $$F(t)=F_0\sin{(\omega t)},$$ is applied to it, where ##F_0## and ##\omega## are constants.

a) Find the velocity and displacement of the proton as functions of time.
b) Show that the proton's distance from the origin grows without bound as time goes on.

Homework Equations



##F(t)=ma(t)## seems to be the only necessary equation.

The Attempt at a Solution



First, we solve for ##a(t)## using ##F(t)=ma(t)##. We have $$ma(t)=F_0\sin{(\omega t)}\rightarrow \frac{dv}{dt}=\frac{F_0}{m}\sin{(\omega t)}.$$ Then we integrate once to find velocity of the particle. We have, $$\int_{0}^{v}dv'=\int_{0}^{t}\frac{F_0}{m}\sin{(\omega t)}\,dt'\rightarrow v(t)=-\frac{F_0}{m\omega}\cos{(\omega t)}.$$ Integrating a second time, we find that the proton's displacement is given by, $$x(t)=-\frac{F_0}{m\omega^2}\sin{(\omega t)}.$$ I'm having trouble showing that the proton's distance from the origin grows without bound, as it seems that the proton should just oscillate between, ##\frac{F_0}{m\omega^2},##, and, ##-\frac{F_0}{m\omega^2}##.
 
Last edited:
Physics news on Phys.org
  • #2
Is the proton initially at rest? There is no such thing specified.
 
  • #3
kuruman said:
Is the proton initially at rest? There is no such thing specified.
Yes, sorry; I've edited that into the question. The proton is indeed at rest initially.
 
  • #4
Something doesn't make sense with your solution. Assuming that ##F_0## is positive, the velocity should be zero at t = 0 and initially increase as time increases. Your expression for ##v(t)## does neither.
 
  • #5
kuruman said:
Something doesn't make sense with your solution. Assuming that ##F_0## is positive, the velocity should be zero at t = 0 and initially increase as time increases. Your expression for ##v(t)## does neither.
But the force is oscillating between ##F_0## and ##-F_0##, so it seems to me that acceleration gained by the proton as a result of the force is also lost as the force oscillates with time. So, the velocity should also be oscillating. I'm not able to find where my math goes wrong either subject to the specified initial conditions. And it's not specified whether or not ##F_0## is negative or positive, only that it is a constant.
 
Last edited:
  • #6
vbrasic said:
But the force is oscillating between ##F_0## and ##-F_0##, so it seems to me that acceleration gained by the proton as a result of the force is also lost as the force oscillates with time. So, the velocity should also be oscillating. I'm not able to find where my math goes wrong either subject to the specified initial conditions. And it's not specified whether or not ##F_0## is negative or positive, only that it is a constant.

This problem clearly specifies simple harmonic motion, so something is wrong.
 
  • #7
vbrasic said:
And it's not specified whether or not ##F_0## is negative or positive, only that it is a constant.
It doesn't matter what sign ##F_0## has. When a force is applied to an object initially at rest, its velocity (an infinitesimally small time interval dt later) will be in the direction of the acceleration (and the force). That's the meaning of dv = a dt.
 
  • #8
PeroK said:
This problem clearly specifies simple harmonic motion, so something is wrong.
Here is my understanding of the physics behind the problem. The particle is initially at rest at the origin. Then, we start applying this force to the particle. The force imparts some momentum to the particle, and it begins to move. But then, the force oscillates so it imparts the same amount of momentum to the particle in the opposite direction of initial motion, so the particle deccelerates, and so on. In this case, would the velocity not be sinusoidally oscillating? I'm having trouble seeing, what about my math subject to these initial conditions is incorrect.
 
  • #9
vbrasic said:
Here is my understanding of the physics behind the problem. The particle is initially at rest at the origin. Then, we start applying this force to the particle. The force imparts some momentum to the particle, and it begins to move. But then, the force oscillates so it imparts the same amount of momentum to the particle in the opposite direction of initial motion, so the particle deccelerates, and so on. In this case, would the velocity not be sinusoidally oscillating? I'm having trouble seeing, what about my math subject to these initial conditions is incorrect.

Yes, with SHM all the time derivatives are sinusoidal, hence the force is sinusoidal. The displacement is sinusoidal and bounded.

Something is wrong with the question.
 
  • #10
PeroK said:
Yes, with SHM all the time derivatives are sinusoidal, hence the force is sinusoidal. The displacement is sinusoidal and bounded.
So, the only issue with this is that the displacement is not unbounded as the question suggests?
 
  • #11
vbrasic said:
So, the only issue with this is that the displacement is not unbounded as the question suggests?

Yes, clearly. It's (bounded) SHM.
 
  • #12
Bounded or unbounded SHM, there is still a problem with OP's posted expression for ##v(t)##. It does not predict that ##v(0) = 0## and that it increases in the direction of the force over the first quarter-period.
 
  • #13
vbrasic said:
Here is my understanding of the physics behind the problem. The particle is initially at rest at the origin. Then, we start applying this force to the particle. The force imparts some momentum to the particle, and it begins to move. But then, the force oscillates so it imparts the same amount of momentum to the particle in the opposite direction of initial motion, so the particle deccelerates, and so on.
Right. Note that this does not return the particle to the origin, but only brings its motion to rest. Essentially, at the beginning of each "cycle" of the force, there should be a constant of integration added representing the offset at the end of the previous cycle.

A way to take this into account without breaking your brain is to turn your integrals into definite integrals, hence:

$$v(t) = \int_0^t \frac{F_o}{m_p} sin(\omega \tau)\;d\tau$$

Then do the same with that result to find the position.
 
  • Like
Likes vbrasic
  • #14
gneill said:
Right. Note that this does not return the particle to the origin, but only brings its motion to rest. Essentially, at the beginning of each "cycle" of the force, there should be a constant of integration added representing the offset at the end of the previous cycle.

A way to take this into account without breaking your brain is to turn your integrals into definite integrals, hence:

$$v(t) = \int_0^t \frac{F_o}{m_p} sin(\omega \tau)\;d\tau$$

Then do the same with that result to find the position.

Nevermind, I caught the problem. I messed up on the bounds of my integral. That is, we have that, $$v(t)=\int_{0}^{t}\frac{F_0}{m_p}\sin{(\omega t')}dt'=-\frac{F_0}{m_p \omega}\cos{(\omega t)}\bigg|_{0}^{t}.$$ I mistakenly assumed that ##\cos{0}=0,## which is obviously not the case. Instead, I should have $$v(t)=-\frac{F_0}{m_p\omega}\cos{(\omega t)}+\frac{F_0}{m_p\omega},$$ which when integrated again, leads to a constant term, which grows without bound. Thanks!
 
  • Like
Likes PeroK
  • #15
vbrasic said:
Nevermind, I caught the problem. I messed up on the bounds of my integral. That is, we have that, $$v(t)=\int_{0}^{t}\frac{F_0}{m_p}\sin{(\omega t')}dt'=-\frac{F_0}{m_p \omega}\cos{(\omega t)}\bigg|_{0}^{t}.$$ I mistakenly assumed that ##\cos{0}=0,## which is obviously not the case. Instead, I should have $$v(t)=-\frac{F_0}{m_p\omega}\cos{(\omega t)}+\frac{F_0}{m_p\omega},$$ which when integrated again, leads to a constant term, which grows without bound. Thanks!

Yes, sorry, I wasn't thinking. It looked like SHM to me, but the initial conditions were wrong. It certainly fooled me.
 
  • #16
vbrasic said:
... which when integrated again, leads to a constant term, which grows without bound. Thanks!
That's more like it.
 

Related to Applying a time-dependent force to a proton

What is a time-dependent force?

A time-dependent force is a force that changes over time. This means that the magnitude and direction of the force experienced by an object will vary as time passes.

How can a time-dependent force be applied to a proton?

A time-dependent force can be applied to a proton through various methods such as an electric field, a magnetic field, or a combination of both.

What effects does a time-dependent force have on a proton?

The effects of a time-dependent force on a proton depend on the specific force and its characteristics. Generally, it can cause the proton to accelerate, change direction, or experience a change in energy.

What are some real-world applications of applying a time-dependent force to a proton?

Some real-world applications include particle accelerators, nuclear fusion reactors, and medical imaging technologies such as MRI machines.

Are there any limitations or challenges in applying a time-dependent force to a proton?

Yes, there are limitations and challenges in applying a time-dependent force to a proton. These include the precision and control required, as well as the potential for damaging the proton or altering its properties.

Similar threads

  • Introductory Physics Homework Help
Replies
17
Views
588
  • Introductory Physics Homework Help
Replies
2
Views
713
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
330
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
19
Views
899
  • Introductory Physics Homework Help
Replies
9
Views
3K
  • Introductory Physics Homework Help
Replies
18
Views
2K
  • Introductory Physics Homework Help
Replies
14
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
735
Back
Top