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:
$\cos(\phi)\sin(\theta)+\sin(\phi)\cos(\theta)=\frac C{A^2+B^2}=\sin(\theta+\phi)$