Integrating acceleration sin wave

[jlv]

This is a really basic calc/physics question.If acceleration is defined as

Acc= Asin(w*t), and I integrate this to get velocity, I get

Vel=(-A/w)*cos(w*t)+C.

If the velocity at t=0 is 0, then C=A/w.

If I then integrate the velocity to get the displacement, I get:

Disp=(-A/w^2)*sin(w*t)+A*t/w+C

If Disp=0 at t=0, then C=0

If I plot this displacement, it is not sinusoidal about x=0 axis. it is always increasing in waves. So, it looks like a sin wave at a 45 degree angle.

Is this right? I was expecting an up and down motion, so a sine wave about x=0.

 

[sunmaggot]

I think maybe you didn't get it wrong. so you used sine function for this. For first T/2, you are accelerated and move forward. Then the next T/2 reduces your velocity into 0. However, your velocity will never go into negative. so I think that displacement keeps increasing is correct. Therefore, it also tells you that your expectation is not correct.

 

[Noctisdark]

You got all right, let's discover why ! The wave is accelerating so it's velocity it's getting bigger and so it's displacing in the positive direction, it's is fine but once the acceleration is becoming negative, the speed is going down but it takes time until it becomes negative and by that fact, even during the decerelation, that charge will be slightly moving in the + direction until it the velocity becomes negative and so on ... so by intuition you expect that the the displacement would be like that !

 

[sunmaggot]

You got all right, let's discover why ! The wave is accelerating so it's velocity it's getting bigger and so it's displacing in the positive direction, it's is fine but once the acceleration is becoming negative, the speed is going down but it takes time until it becomes negative and by that fact, even during the decerelation, that charge will be slightly moving in the + direction until it the velocity becomes negative and so on ... so by intuition you exoect that the the displacement would be like that !

For this case, velocity never goes to negative. It only fluctuates between maximum velocity (it is positive) and 0 velocity. This can be easily understood by the a-t graph. Let me give you an example, if the acceleration is cosine function, then the net displacement for one period will be 0.

 

[Noctisdark]

For this case, velocity never goes to negative. It only fluctuates between maximum velocity (it is positive) and 0 velocity. This can be easily understood by the a-t graph. Let me give you an example, if the acceleration is cosine function, then the net displacement for one period will be 0.

I've missed that out, Thanks,
Since v is allways >0 then the displacement keep getting bigger, and you should expect the result you've got !

 

[jlv]

Thank you all for your helpful responses. I agree, with V>0 always, the integral will be positive and increasing. I was thinking more in terms of V above and below x=0. But, that would only happen if I set the appropriate initial velocity.

 

[nasu]

If you expected to get the standard harmonic motion, think about the fact that for this kind of motion you never have all three variables (x,v,a) going through zero at the same time. Your initial conditions are not compatible with such a motion.

 

[theodoros.mihos]

This is an classic problem with initial conditions.
At t=0 you say v=0 and x=0. Combine Newton's law with Hook's law we have:
$$ a(t) = \frac{F}{m} = -\frac{k}{m}x(t) = -\omega^2x(t) $$
So, if a(t) have only sinusoidal part, the same must be with x(t). Complete solution is:
$$ \frac{d^2x}{dt^2} = a_0\sin(\omega{t}) \Rightarrow \frac{dx}{dt}= v(t) = c_1 -\frac{a_0}{\omega}\cos(\omega{t}) $$
$$ v(0)=-v_0 = c_1-\frac{a_0}{\omega} \Rightarrow c_1 = \frac{a_0}{\omega}-v_0 $$
$$ x(t) = c_2 + c_1t -\frac{a_0}{\omega^2}\sin(\omega{t}) \Rightarrow c_2=0 $$
So, general solution is:
$$ x(t) =\left(\frac{a_0}{\omega}-v_0\right)t -\frac{a_0}{\omega^2}\sin(\omega{t}) ,\, v(t) = \frac{a_0}{\omega}-v_0 - \frac{a_0}{\omega}\cos(\omega{t}) $$
But the original equation demands x(t) is only sinusoidal so:
$$ \frac{a_0}{\omega}-v_0=0 \Leftrightarrow v_0=\frac{a_0}{\omega} $$
and general solution been:
$$ x(t) = -\frac{a_0}{\omega^2}\sin(\omega{t}) ,\, v(t) = - \frac{a_0}{\omega}\cos(\omega{t}) $$
which is compartible with the condition
$$ v_0=\frac{a_0}{\omega} $$

Always you can see the general solution with the additional time term as the solution for every frame which move with v according to our frame.

 

[sunmaggot]

This is an classic problem with initial conditions.
At t=0 you say v=0 and x=0. So, this problem have the trivial zero solution. Combine Newton's law with Hook's law we have:
$$ a(t) = \frac{F}{m} = -\frac{k}{m}x(t) = -\omega^2x(t) $$
So, if a(t) have only sinusoidal part, the same must be with x(t). Complete solution is:
$$ \frac{d^2x}{dt^2} = a_0\sin(\omega{t}) \Rightarrow \frac{dx}{dt}= v(t) = c_1 -\frac{a_0}{\omega}\cos(\omega{t}) $$
$$ v(0)=-v_0 = c_1-\frac{a_0}{\omega} \Rightarrow c_1 = \frac{a_0}{\omega}-v_0 $$
$$ x(t) = c_2 + c_1t -\frac{a_0}{\omega^2}\sin(\omega{t}) \Rightarrow c_2=0 $$
So, general solution is:
$$ x(t) =\left(\frac{a_0}{\omega}-v_0\right)t -\frac{a_0}{\omega^2}\sin(\omega{t}) ,\, v(t) = \frac{a_0}{\omega}-v_0 - \frac{a_0}{\omega}\cos(\omega{t}) $$
But the original equation demands x(t) is only sinusoidal so:
$$ \frac{a_0}{\omega}-v_0=0 \Leftrightarrow v_0=\frac{a_0}{\omega} $$
and general solution been:
$$ x(t) = -\frac{a_0}{\omega^2}\sin(\omega{t}) ,\, v(t) = - \frac{a_0}{\omega}\cos(\omega{t}) $$

I don't understand why you linked simple harmonic motion with this problem. First, for simple harmonic motion, acceleration is a cosine function. The condition for the above problem is a sine function, which is different. Second, I think, judging from a-t graph, the displacement will actually keep increasing. For a sine curve, the first T/2, you get positive area, which means positive velocity. After the next T/2, you get 0 area because the positive and negative areas have same absolute area. so what happens is that, the area will never go to negative. so displacement will keep increasing because V is always bigger or equal to 0.

There is a possibility that I misunderstood your general solution, because right now I still don't get what you concluded from those formula.

 

[theodoros.mihos]

The difference between sin and cos is a phase=pi/2 but ok, you are right. There is not an harmonic problem so it has another solution:
$$ v(t)= c - \frac{a_0}{\omega}\cos(\omega{t}),\, x(t) = ct -\frac{a_0}{\omega^2}\sin(\omega{t}) $$
There is a motion with two terms: the 1st represent a normal constant velocity c and the 2nd an additional harmonic term. It is normal, like frog movement. The value of c must be the maximum velocity for t=T/4. By energy conservation:
$$ W = \int_{t=0}^{t=T/4}Fdx = \int_{t=0}^{t=T/4}ma(t)\,dx = \frac{1}{2}mc^2 \Rightarrow \int_{0}^{T/4}a(t)v(t)dt = \frac{1}{2}c^2 \Rightarrow $$
$$ c^2 = 2\int_0^{T/4}\frac{a_0^2}{\omega}\sin(\omega{t})\cos(\omega{t})dt = 2\frac{a_0^2}{\omega^2}\int_0^{1}u\,du = \frac{a_0^2}{\omega^2} \Rightarrow c = \frac{a_0}{\omega} $$

The problem is an harmonic oscillator, as can see an observer on frame with velocity c :-)

 

[sunmaggot]

The difference between sin and cos is a phase=pi/2 but ok, you are right. There is not an harmonic problem so it has another solution:
$$ v(t)= c - \frac{a_0}{\omega}\cos(\omega{t}),\, x(t) = ct -\frac{a_0}{\omega^2}\sin(\omega{t}) $$
There is a motion with two terms: the 1st represent a normal constant velocity c and the 2nd an additional harmonic term. It is normal, like frog movement. The value of c must be the maximum velocity for t=T/4. By energy conservation:
$$ W = \int_{t=0}^{t=T/4}Fdx = \int_{t=0}^{t=T/4}ma(t)\,dx = \frac{1}{2}mc^2 \Rightarrow \int_{0}^{T/4}a(t)v(t)dt = \frac{1}{2}c^2 \Rightarrow $$
$$ c^2 = 2\int_0^{T/4}\frac{a_0^2}{\omega}\sin(\omega{t})\cos(\omega{t})dt = 2\frac{a_0^2}{\omega^2}\int_0^{1}u\,du = \frac{a_0^2}{\omega^2} \Rightarrow c = \frac{a_0}{\omega} $$

The problem is an harmonic oscillator, as can see an observer on frame with velocity c :-)

did the above question set a frame with velocity? you are not answering the problem, can I say so? the problem in your calculation, you are using different frames in same event. First, assume the problem sets frame as stationary, the frame observes a sine function of acceleration. If the frame becomes a moving one, the acceleration function will be different. You use the acceleration function obtained from stationary frame, and use a moving frame to evaluate the x(t), this is obviously not correct. Therefore, there is no simple harmonic motion in the above problem.

 

[theodoros.mihos]

Once again:
$$ v(t)= \frac{a_0}{\omega} - \frac{a_0}{\omega}\cos(\omega{t}),\, x(t) = \frac{a_0}{\omega}t -\frac{a_0}{\omega^2}\sin(\omega{t}) $$
but I think this problem is not for a student steel belives that the velocity is not relative.

 

[sunmaggot]

Once again:
$$ v(t)= \frac{a_0}{\omega} - \frac{a_0}{\omega}\cos(\omega{t}),\, x(t) = \frac{a_0}{\omega}t -\frac{a_0}{\omega^2}\sin(\omega{t}) $$
but I think this problem is not for a student steel belives that the velocity is not relative.

but this problem is also not for a student who does not know what is meant by "use one frame for one whole event"

 

[theodoros.mihos]

good luck my friend :-)

 

[sunmaggot]

good luck my friend :-)

v = 5 m/s
s = vt = 5t
what is the displacement after one second?
0, because for a frame with v = 5 m/s, displacement is 0. Is this a proper answer?
What do you mean by good luck?

 

[theodoros.mihos]

After 1 period T=2π/ω, for the "unmoving" frame is:
$$ v=0,\, x=2\pi\frac{a_0}{\omega^2} $$
from its solution $$ v(t) = \frac{a_0}{\omega}[1-\cos(\omega{t})],\,x(t)=\frac{a_0}{\omega}t-\frac{a_0}{\omega^2}\sin(\omega{t}) $$
and for a frame moving with v=a₀/ω relative to "unmoving":
$$ v=-\frac{a_0}{\omega},\, x=0 $$
from its solution $$ v(t) = - \frac{a_0}{\omega}\cos(\omega{t}),\,x(t)=-\frac{a_0}{\omega^2}\sin(\omega{t}) $$
Take that you want. Acceleration, the only physical condition in this problem, is the same for all frames are moving with constant velocity.
I give you two equivalent solution and you can't see one. For this I say: good luck.