Frequency of Vibration of Two Masses Joined by a Spring

[flamespirit919]

Homework Statement

Two masses $m_1$ and $m_2$ are joined by a spring of spring constant $k$. Show that the frequency of vibration of these masses along the line connecting them is:
$$\omega =\sqrt{\frac{k(m1+m2)}{m1m2}}$$

Homework Equations

$x(t)=Acos(\omega t)$
$\omega =\sqrt{\frac{k}{m}}$
$m\frac{d^2x}{dt^2}=-kx$

The Attempt at a Solution

So I have that the distance traveled by $m_1$ can be represented by the function $x_1(t)=Acos(\omega t)$ and similarly for the distance traveled by $m_2$ is $x_2(t)=Bcos(\omega t)$. The force the spring exerts on these two masses is $−kx_n(t)=m_n\frac{d^2x_n}{dt^2}$. But I have no idea how to relate these functions. Plugging in the values I get the two functions $$-kAcos(\omega t)=-m_1A\omega ^2cos(\omega t)$$ $$-kBcos(\omega t)=-m_2B\omega ^2cos(\omega t)$$

Simplifying I get $$k=m_1\omega ^2$$ $$k=m_2\omega^2$$
But, then I don't know what to do after this or if I am even in the right direction.

 

[TSny]

The force the spring exerts on these two masses is $−kx_n(t)=m_n\frac{d^2x_n}{dt^2}$.

You are taking the force on m₁ to be -kx₁. But the elongation of the spring depends on both x₁ and x₂, not just x₁.

 

[kuruman]

Suppose you say that $m_1$ is displaced to the right by amount $x_1$ from equilibrium and that $m_2$ is also displaced to the right by amount $x_2$ from equilibrium. What is the amount by which the spring is stretched or compressed from equilibrium? Can you write a differential equation of motion for each mass (Newton's 2nd law)?

 

[flamespirit919]

Suppose you say that $m_1$ is displaced to the right by amount $x_1$ from equilibrium and that $m_2$ is also displaced to the right by amount $x_2$ from equilibrium. What is the amount by which the spring is stretched or compressed from equilibrium? Can you write a differential equation of motion for each mass (Newton's 2nd law)?

So would it be $$-k(x_1+x_2)=(m_1+m_2)(\frac{d^2x_1}{dt^2}+\frac{d^2x_2}{dt^2})$$$$-k(Acos(\omega t)+Bcos(\omega t)=(m_1+m_2)(-A\omega ^2cos(\omega t)-B\omega ^2cos(\omega t)$$ Which simplifies to $$k=\omega ^2 (m_1+m_2)$$ But, what can I do with this? Or did I go about this the wrong way?

 

[kuruman]

So would it be $$
-k(x_1+x_2)=(m_1+m_2)(\frac{d^2x_1}{dt^2}+\frac{d^2x_2}{dt^2})$$ ...

No. If, say, $x_1$ = 3 cm and $x_2 = 5$ cm, would the spring be stretched by 8 cm? Draw a picture to scale, first with the spring in equilibrium and then redraw it with one end displaced by 3 units to the right and the other by 5 units also to the right. By how many units is the spring stretched? Knowing the stretching of the spring from equilibrium, gives you the force on each mass. Write separate differential equations for each mass.

 

[flamespirit919]

No. If, say, $x_1$ = 3 cm and $x_2 = 5$ cm, would the spring be stretched by 8 cm? Draw a picture to scale, first with the spring in equilibrium and then redraw it with one end displaced by 3 units to the right and the other by 5 units also to the right. By how many units is the spring stretched? Knowing the stretching of the spring from equilibrium, gives you the force on each mass. Write separate differential equations for each mass.

The spring would be stretched by 2 units. Would the equations then be $$-k(x_2-x_1)=m_1\frac{d^2x_1}{dt^2}$$$$-k(x_2-x_1)=m_2\frac{d^2x_2}{dt^2}$$

 

[TSny]

The spring would be stretched by 2 units. Would the equations then be $$-k(x_2-x_1)=m_1\frac{d^2x_1}{dt^2}$$$$-k(x_2-x_1)=m_2\frac{d^2x_2}{dt^2}$$

Almost. Does the force of the spring on m₁ have the same direction as the force on m₂?

 

[flamespirit919]

Almost. Does the force of the spring on m₁ have the same direction as the force on m₂?

No, so would it be $$k(x_2-x_1)=m_1\frac{d^2x_1}{dt^2}$$$$-k(x_2-x_1)=m_2\frac{d^2x_2}{dt^2}$$ Assuming that $m_1$ is on the left and $m_2$ is on the right.

 

[TSny]

Looks like some typos in your latest equations on the left side. Please check to see if you typed them the way you wanted.

 

[flamespirit919]

Looks like some typos in your latest equations on the left side. Please check to see if you typed them the way you wanted.

Whoops. They should be fixed now.

 

[TSny]

Looks good. Can you see a way to solve these equations?

 

[flamespirit919]

Looks good. Can you see a way to solve these equations?

I tried setting them equal to each other $$-m_1A\omega ^2cos(\omega t)=m_2B\omega ^2cos(\omega t)$$ But I end up with $$-m_1A=m_2B$$ I don't know what I can do after this.

 

[TSny]

I tried setting them equal to each other $$-m_1A\omega ^2cos(\omega t)=m_2B\omega ^2cos(\omega t)$$ But I end up with $$-m_1A=m_2B$$

OK. This gives you a relation between A and B in the assumptions x₁ = Acosωt and x₂ = Bcosωt. Try to combine this result with the first differential equation k(x₂ - x₁) = m₁d²x₁/dt².

 

[flamespirit919]

OK. This gives you a relation between A and B in the assumptions x₁ = Acosωt and x₂ = Bcosωt. Try to combine this result with the first differential equation k(x₂ - x₁) = m₁d²x₁/dt².

Thank you! I wasn't sure which equation to plug the values into. Here's what I did $$k(Bcos(\omega t)-Acos(\omega t))=-m_1A\omega ^2cos(wt)$$$$k(Bcos(\omega t)+\frac{m_2B}{m_1}cos(\omega t)=m_2B\omega ^2cos(\omega t)$$$$kBcos(\omega t)(1+\frac{m_2}{m_1})=m_2B\omega ^2cos(\omega t)$$$$k(1+\frac{m_2}{m_1})=m_2\omega ^2$$$$k(m_1+m_2)=m_1m_2\omega ^2$$$$\omega ^2=\frac{k(m_1+m_2)}{m_1m_2}$$$$\omega =\sqrt{\frac{k(m_1+m_2)}{m_1m_2}}$$Did I do all my math right?

 

[zwierz]

Let $x_1<x_2$ be coordinates of masses $m_1,m_2$ respectively. Then
$m_2\ddot x_2=-(x_2-x_1-l)k,\quad m_1\ddot x_1=(x_2-x_1-l)k$, here $l$ is the length of relaxed spring. Divide each equation by corresponding mass and subtract the second equation from the first one
$\ddot\xi=-(k/m_2+k/m_1)\xi+const,\quad \xi=x_2-x_1$ so that $\omega^2= k/m_2+k/m_1$

 

[TSny]

Did I do all my math right?

Yes, I believe so.

Note that in your approach, you assumed x₁ and x₂ have time dependences of cosωt. That might be OK.

@zwierz shows how to get a differential equation that directly gives you the frequency without having to make any prior assumptions about the nature of the motion.