Derive angular frequency for mass spring system

In summary, the individual masses are connected in series and the extension of the spring is expressed in terms of x1/2, x2-x1, and l. The relation between x and x_2-x_1 is solved by substituting between x = x2-x1 and x = x.
  • #1
so_gr_lo
69
10
Homework Statement
Two masses m1 and m2 on the x axis are connected by a spring. The spring has stiffness s, length l and extension x. m2 is at position x2 and m1 at position x1. The equations of the motion are

m1d^2x1/dt^2 = sx and m2d^2x2/dt^2 = -sx

Combine these to show that the angular frequency is w = sqrt(s/M)

Where M = m1m2/ m1 + m2 (the reduced mass)
Relevant Equations
m1d^2x1/dt^2 = sx and m2d^2x2/dt^2 = -sx

w = sqrt(s/M)

M = m1m2/ m1 + m2
tried writing the x position as

x = Acos(wt) (ignoring the phase)

so that d2x / dt2 = -w2x

Substituting that into the individual motion equations would get the required result for the individual masses, but I am not sure how to combine the equations to get the reduced mass
 
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  • #2
Can you express the extension ##x## in terms of ##x_1##, ##x_2##, and ##l##?
 
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  • #3
Are the masses connected in series or to either ends of the spring?
 
  • #4
Expresign extension in term of x1/2 gives

x2-x1 = x

Which could be substituted into each of the motion equatiosn but I'm not sure how that helps
 
  • #5
VVS2000 said:
Are the masses connected in series or to either ends of the spring?
Yes the masses are at the ends
 
  • #6
so_gr_lo said:
Expresign extension in term of x1/2 gives

x2-x1 = x

Which could be substituted into each of the motion equatiosn but I'm not sure how that helps
##x_2 - x_1## is the distance from one end of the spring to the other end of the spring. This distance will include the natural length of the spring ##l## and the extension ##x##.

Once you express ##x## in terms of ##x_1## and ##x_2##, you can see how ##\ddot x## is related to ##\ddot x_2## and ##\ddot x_1##.
 
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  • #7
Is d2x/dt2 = d2x2/dt2 - d2x1/dt2 ?
 
  • #8
so_gr_lo said:
Is d2x/dt2 = d2x2/dt2 - d2x1/dt2 ?
Yes. But I can't tell if you arrived at this correctly. What equation did you use for the relation between ##x## and ##x_2 - x_1##?
 
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  • #9
Here is a figure to help you out with setting the equations. When the spring is unstretched, the masses are at the dotted lines, i.e. ##x_1-x_1=L##.
Twomasseswithspring.png
 
  • #10
TSny said:
Yes. But I can't tell if you arrived at this correctly. What equation did you use for the relation between ##x## and ##x_2 - x_1##?
I used between x = x2-x1
 
  • #11
so_gr_lo said:
I used between x = x2-x1
This is not correct.
##x## represents the amount of stretch of the spring from its unstretched length ##l##. So, if ##x = 0##, then ##x_2 - x_1## = ##l##. Your equation doesn't satisfy this condition.
 
  • #12
So x = x2-x1-l
 
  • #13
so_gr_lo said:
So x = x2-x1-l
Yes, that's the correct relation.
 
  • #14
Okay, but if I substitute that into the equations I get

m1(d2x2/dt2) - d2x1/dt2) = sx

and similar for m2, how does this help with combining them?
 
  • #15
so_gr_lo said:
Okay, but if I substitute that into the equations I get

m1(d2x2/dt2) - d2x1/dt2) = sx

This is not correct. You already had the correct equations of motion for ##m_1## and ##m_2## in the "Relevant Equations" section of your first post.

You also have ##\ddot x = \ddot x_2 - \ddot x_1##. Use the equations of motion to express the right side in terms of ##s##, ##x##, and the masses.
 

FAQ: Derive angular frequency for mass spring system

What is the formula for deriving angular frequency for a mass spring system?

The formula for deriving angular frequency for a mass spring system is ω = √(k/m), where ω represents the angular frequency, k represents the spring constant, and m represents the mass of the object.

How is angular frequency related to the natural frequency of a mass spring system?

Angular frequency and natural frequency are directly proportional in a mass spring system. The natural frequency is equal to the angular frequency divided by 2π.

Can the angular frequency of a mass spring system be negative?

No, the angular frequency of a mass spring system cannot be negative. It is always a positive value, representing the rate at which the object oscillates around its equilibrium position.

How does changing the mass or spring constant affect the angular frequency of a mass spring system?

If the mass of the object increases, the angular frequency decreases. Conversely, if the spring constant increases, the angular frequency also increases. This relationship is described by the formula ω = √(k/m).

What are the units of angular frequency for a mass spring system?

The units of angular frequency for a mass spring system are radians per second (rad/s). This represents the amount of rotation per unit time.

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