Solve for Period of Oscillation of 2 Springs with Mass Attached

In summary: Then:F = -kx1 = -kx2F = -kx3 = -kx4Where x1, x2, x3, and x4 are the displacements of the four springs, respectively. Since x1 = x2 and x3 = x4, we can rewrite the equations as:F = -2kx1F = -2kx3Now, considering the forces on the mass, we have:F = -kx1 - kx3 = maSubstituting the values of F from the previous equations, we get:-2kx1 = -kx1 - kx3 = maSolving for a, we
  • #1
oneplusone
127
2

Homework Statement


You have two springs with spring constants k_1 = 10 and k_2 = 20 vertically attached to a wall, and a mass of mass 3.00kg is hung from it. Find the period of oscillation.


----------------
||
|| <-- first spring, k_1 = 10
||
--
||
|| <== second spring, k_2 = 20
||
{} <== mass 3.00kg

Homework Equations



F=-kx
T = 2pi/omega


The Attempt at a Solution




Consider the second spring. We have:
[itex] F=-kx \implies 3(-9.8) = -(20)(x) \implies x = 1.47 [/itex]

Now the second spring undergoes a displacement of:

[itex] F=-kx \implies 3(-9.8) = -(10)x \implies x = 2.94 [/itex].

So k effective is:

[itex] F = -k (x_1+x_2) \implies 3(9.8) = k(2.94+1.47) \implies k = 4.41[/itex]

From here, i just found [itex]\omega[/itex] by using [itex]\omega =\sqrt{k}{m}[/itex]. and the period was easy from there.

Is the first part of my solution correct? I am having trouble visualizing it.
 
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  • #2
oneplusone said:
Consider the second spring. We have:
[itex] F=-kx \implies 3(-9.8) = -(20)(x) \implies x = 1.47 [/itex]

Now the second spring undergoes a displacement of:

[itex] F=-kx \implies 3(-9.8) = -(10)x \implies x = 2.94 [/itex].
So far so good.

So k effective is:

[itex] F = -k (x_1+x_2) \implies 3(9.8) = k(2.94+1.47) \implies k = 4.41[/itex]
4.41 is the total displacement, not the spring constant. Solve for the effective spring constant.
 
  • #3
oops i meant 6.666. Thank you.

A follow up question to this has the same springs, except oriented like this:

-------------------
|| ||
|| ||
|| ||
|| ||
------
| {} |
------With the springs parallel. How would you solve this? I am still not convinced that there is a solution (to me, the "block" will be tilted).
 
  • #4
oneplusone said:
With the springs parallel. How would you solve this? I am still not convinced that there is a solution (to me, the "block" will be tilted).
Assume the system is constrained so that the springs get the same displacement.
 
  • #5


Yes, your solution is correct. By using the equation F=-kx, you have correctly found the displacements of both springs. The effective spring constant, k effective, is also calculated correctly by considering both springs in series. This value is then used to find the angular frequency, ω, and the period of oscillation can be easily calculated from there. Your solution is also clearly presented and easy to follow. Good job!
 

1. What is the formula for calculating the period of oscillation for 2 springs with mass attached?

The formula for calculating the period of oscillation for 2 springs with mass attached is T = 2π√(m/k), where T is the period, m is the mass attached to the springs, and k is the spring constant.

2. How do the masses and spring constants affect the period of oscillation?

The period of oscillation is directly proportional to the square root of the mass attached and inversely proportional to the square root of the spring constant. This means that as the mass increases, the period will increase, and as the spring constant increases, the period will decrease.

3. Can the period of oscillation for 2 springs with mass attached be affected by external factors?

Yes, the period of oscillation can be affected by external factors such as friction, air resistance, and damping forces. These external factors can alter the oscillation of the springs and affect the calculated period.

4. How does the arrangement of the springs affect the period of oscillation?

The arrangement of the springs can affect the period of oscillation by changing the effective spring constant. For example, if the springs are arranged in series, the effective spring constant will be lower and the period will be longer compared to when the springs are arranged in parallel.

5. Are there any real-life applications of calculating the period of oscillation for 2 springs with mass attached?

Yes, there are many real-life applications of this calculation, such as in pendulum clocks, suspension bridges, and shock absorbers in cars. It is also used in the study of mechanical systems and vibrations in engineering and physics.

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