Position and acceleration in simple harmonic motion given velocity

[zenterix]

Homework Statement

This problem is from Apostol's Calculus, Volume I.

A particle undergoes simple harmonic motion. Initially its displacement is 1, its velocity is 1, and its acceleration is -12.

Compute its displacement and acceleration when the velocity is ##\sqrt{8}##.

Relevant Equations

##F=ma##

A spring attached to a mass undergoes simple harmonic motion.

From Newton's second law we have ##ma=-qx## where ##q## is the spring constant.

$$x''+\frac{q}{m}x=0$$

A second order equation with constant coefficients.

The characteristic equation is ##r^2+\frac{q}{m}=0##. The roots are complex.

The discriminant is ##\Delta = -\frac{4q}{m}<0##.

Let ##k=\frac{1}{2}\sqrt{-\Delta}=\frac{1}{2}\sqrt{\frac{4q}{m}}##

Then the general solution to our differential equation is

$$x(t)=c_1\cos{kt}+c_2\sin{kt}$$

Since ##x(0)=1## then ##c_1=1##.

Then

$$v(t)=-k\sin{kt}+c_2k\cos{kt}$$

Since ##v(0)=1## then ##c_2=\frac{1}{k}##.

Then,

$$a(t)=-\frac{k}{m}x(t)$$

Since ##a(0)=-\frac{k}{m}=-12## then we can solve for ##q=\frac{24^2m}{4}##.

Thus, at this point we have

$$x(t)=\cos{kt}+\frac{1}{k}\sin{kt}$$
$$v(t)=-k\sin{kt}+\cos{kt}$$
$$a(t)=-\frac{k}{m}\cos{kt}-\frac{1}{m}\sin{kt}$$

Suppose ##v(t)=2\sqrt{2}##.

If we can solve for the corresponding ##t## then we can plug this ##t## into ##x(t)## and ##a(t)##.

But how?

 

[haruspex]

You don't need to involve mass, force or spring constant. Neither do you need to solve any DE.
Just start with the trig function for x(t). Find ##c_1, c_2## as you did, but then apply the same method to acceleration to find k.

 

[zenterix]

You don't need to involve mass, force or spring constant. Neither do you need to solve any DE.
Just start with the trig function for x(t). Find ##c_1, c_2## as you did, but then apply the same method to acceleration to find k.

If we differentiate ##v(t)=-k\sin{kt}+\cos{kt}## to obtain ##a(t)=-k^2\cos{kt}-k\sin{kt}##, use ##a(0)=-12## and solve for ##k## we find ##k=2\sqrt{3}##.

Thus, ##v(t)=-2\sqrt{3}\sin{(2\sqrt{3}t)}+\cos{(2\sqrt{3}t)}##.

If we equate this to ##2\sqrt{2}## we still need to solve for ##t##.

 

[haruspex]

If we differentiate ##v(t)=-k\sin{kt}+\cos{kt}## to obtain ##a(t)##, use ##a(0)=-12## and solve for ##k## we find ##k=2\sqrt{3}##.

Thus, ##v(t)=-2\sqrt{3}\sin{(2\sqrt{3}t)}+\cos{(2\sqrt{3}t)}##.

If we equate this to ##2\sqrt{2}## we still need to solve for ##t##.

There are several ways to proceed.
You can move one of the trig functions across and square.
Or get the RHS into the form ##\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)##.
Or start over with the SHM form ##A\sin(\omega t+\phi)##.

 

[Orodruin]

I suggest doing this without ever involving trig functions (ie, you do not need the actual explicit form of the solution). Note that, apart from ##x’’ + \omega^2 x = 0##, this has a first integral ##x’^2 + \omega^2 x^2 = E## for some constant ##E##.

 

[erobz]

I suggest doing this without ever involving trig functions (ie, you do not need the actual explicit form of the solution). Note that, apart from ##x’’ + \omega^2 x = 0##, this has a first integral ##x’^2 + \omega^2 x^2 = E## for some constant ##E##.

Should it be?

$$ \cancel{2 \dot x ^2 + \omega^2 x^2 = E} $$

Never Mind!

Either way this is great simplification from messing around with ##x = A \sin( \omega t + \phi )## ( I never would have thought of it)

 

[Orodruin]

I never would have thought of it

Let’s just say it is not my first rodeo

 

[haruspex]

this is great simplification from messing around with ##x = A \sin( \omega t + \phi )## ( I never would have thought of it)

And I suspect the author intends this approach. Had the question been to find displacement / acceleration / velocity at a particular time the shortcut would not have been available.

 

[Orodruin]

And I suspect the author intends this approach. Had the question been to find displacement / acceleration / velocity at a particular time the shortcut would not have been available.

Among somethings I picked up during years of teaching GR: If you can avoid the nitty gritty details of finding the explicit solution by using constants of motion, then you better do so.

 

[zenterix]

I suggest doing this without ever involving trig functions (ie, you do not need the actual explicit form of the solution). Note that, apart from ##x’’ + \omega^2 x = 0##, this has a first integral ##x’^2 + \omega^2 x^2 = E## for some constant ##E##.

How does ##\int (x''(t)+\omega^2x(t))dt## give us ##x'^2(t)+\omega^2x^2(t)##?

 

[haruspex]

How does ##\int (x''(t)+\omega^2x(t))dt## give us ##x'^2(t)+\omega^2x^2(t)##?

Not directly. It's a standard trick: multiply the equation by ##\dot x## first. Then all terms are integrable.

 

[erobz]

How does ##\int (x''(t)+\omega^2x(t))dt## give us ##x'^2(t)+\omega^2x^2(t)##?

I'm not familiar with the trick @haruspex proposes, but I began by changing the independent variable in the derivative to ##x##, by application of the chain rule.

 

[zenterix]

$$x''+\omega^2x=0\tag{1}$$
$$x'x''+\omega^2x'x=0\tag{2}$$
$$\frac{x'^2}{2}+\omega^2\frac{x^2}{2}=C_1\tag{3}$$
$$x'^2+\omega^2x^2=C\tag{4}$$

From (1), since ##x''(0)=-12##

$$-12+\omega^2\cdot 1=0\tag{5}$$
$$\omega^2=12\tag{6}$$

From (4), since ##x'(0)=x(0)=1##

$$1+\omega^2\cdot 1=C\tag{7}$$

$$C=1+\omega^2\tag{8}$$

Again from (4), since at some time ##t## we have ##x'(t)=\sqrt{8}## then

$$8+\omega^2x^2(t)=1+\omega^2\tag{9}$$

$$x^2(t)=\frac{-7+\omega^2}{\omega^2}=\frac{-7+12}{12}=\frac{5}{12}\tag{10}$$

$$x(t)=\sqrt{\frac{5}{12}}\tag{11}$$

Then from (1) again

$$x''(t)=-\omega^2x(t)=-12\sqrt{\frac{5}{12}}\tag{12}$$

Unfortunately, the solution manual has

$$x(t)=\frac{1}{3}\sqrt{6}\tag{13}$$

$$x''(t)=-12x(t)=-4\sqrt{6}\tag{14}$$

 

[zenterix]

I redid the calculations using the sinusoidal expressions and solved them with Maple. They agree with the solution proposed here in this forum. The solution manual seems to be incorrect.

$$x''+\omega^2x=0\tag{1}$$

The general solution to this equation is

$$x(t)=c_1\cos{\omega t}+c_2\sin{\omega t}\tag{2}$$

$$x(0)=c_1=1\tag{3}$$

$$v(t)=-\omega \sin{\omega t}+c_2\omega\cos{\omega t}\tag{4}$$

$$v(0)=c_2\omega=1 \implies c_2=\frac{1}{\omega}\tag{5}$$

Thus,

$$v(t)=-\omega\sin{\omega t}+\cos{\omega t}\tag{6}$$

$$a(t)=-\omega^2\cos{\omega t}-\omega\sin{\omega t}\tag{7}$$

$$a(0)=-\omega^2=-12\implies \omega^2=12\implies \omega=2\sqrt{3}\tag{8}$$

Thus,

$$v(t)=-2\sqrt{3}\sin{2\sqrt{3}t}+\cos{2\sqrt{3} t}\tag{9}$$

and we want to find the time ##t## at which ##v(t)=\sqrt{8}=2\sqrt{2}##.

The times comes out to be ##-0.179##.

If we sub into ##x(t)## then we get ##0.645##, which is ##\sqrt{\frac{5}{12}}##.

If we sub into ##a(t)## then we get ##-7.74##, which is ##-12\sqrt{\frac{5}{12}}##.

 

[Orodruin]

$$x(t)=\sqrt{\frac{5}{12}}\tag{11}$$

Your (11) is missing one solution, but otherwise ok.

 

[haruspex]

I'm not familiar with the trick @haruspex proposes, but I began by changing the independent variable in the derivative to ##x##, by application of the chain rule.

Yes, the two methods are effectively the same:
##\ddot x+\omega^2 x=0##
##2\ddot x\dot x+2\omega^2 x\dot x=0##
Integrating wrt t:
##\dot x^2+\omega^2x^2=constant##

Or using ##\ddot x= \frac{d\dot x}{dt}=\frac{d\dot x}{dx}\frac{dx}{dt}=v'v##
##v'v+\omega^2 x=0##
Integrating wrt x:
##v^2+\omega^2 x^2=constant##

 

[haruspex]

Just to illustrate for future use…

There are several ways to proceed.
You can move one of the trig functions across and square.
Or get the RHS into the form ##\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)##.

##A\sin(\theta)+B\cos(\theta)=C##
1.
##A\sin(\theta)=C-B\cos(\theta)##
##A^2\sin^2(\theta)=(C-B\cos(\theta))^2=A^2(1-\cos^2(\theta))##
a quadratic in cos theta
2.
Define ##\phi## by ##\cos(\phi)=\frac A{A^2+B^2}## etc:
##\cos(\phi)\sin(\theta)+\sin(\phi)\cos(\theta)=\frac C{A^2+B^2}=\sin(\theta+\phi)##