An object attached to three springs

[jd12345]

Homework Statement

An object of mass m is attached to three springs each of springs contant k. If the object is pushed slightly towards one of the springs find the time period of the oscillation

The springs are at equal angles from each other - which is 120 degrees.
The other ends of springs are attached to walls


2. The attempt at a solution
I have no idea how to start this question which contains three springs!

By the way, luckily i got the answer which is 2∏√(2m/3k)
please give the solution along with concept - i need to know how to solve problems where object is attached to more than one spring

 

[tiny-tim]

hi jd12345!

find the strength of the restoring force as a function of displacement …

if that's approximately proportional to the displacement (for small values), then the motion will be harmonic, and you can easily find the period

 

[jd12345]

hi jd12345!

find the strength of the restoring force as a function of displacement …

if that's approximately proportional to the displacement (for small values), then the motion will be harmonic, and you can easily find the period

Its not that easy. Displacement of one of the spring agaisnt which it is pushed be x.
IT provides an upwards force kx. But its hard to find elongation of other two springs

 

[tiny-tim]

hi jd12345!

(just got up :zzz: …)

… But its hard to find elongation of other two springs

hard?

Pythagoras could have done it!​

 

[asteriatic]

I am happy to see that you were able to find the solution to this problem. To understand the concept behind this, let's break it down step by step.

First, let's consider the situation where the object is only attached to one spring. In this case, the object will oscillate back and forth with a certain frequency, which we call the natural frequency. This frequency is determined by the mass of the object and the spring constant, and it can be calculated using the formula f = 1/2∏√(k/m). This frequency tells us how many times the object will oscillate in one second.

Now, let's add another spring to the equation. When the object is attached to two springs, it will still oscillate back and forth, but the natural frequency will be different. This is because the springs are now working together to support the object, and their combined stiffness is greater than that of just one spring. This means that the object will oscillate with a higher frequency, and this frequency can be calculated using the same formula as before, but with the total spring constant (k1 + k2) and the same mass (m).

Finally, let's add the third spring. With three springs, the object will oscillate with an even higher frequency, as the combined stiffness of the three springs is even greater. In this case, the natural frequency can be calculated using the same formula, but with the total spring constant (k1 + k2 + k3) and the same mass (m).

Now, let's apply this concept to the given problem. Since the three springs are at equal angles from each other, we can assume that they are all working together to support the object. This means that the total spring constant is the sum of the individual spring constants, which is 3k. Plugging this into the formula for natural frequency, we get f = 1/2∏√(3k/m).

However, we are not looking for the frequency, but rather the time period of the oscillation. The time period is simply the inverse of the frequency, so we can rewrite the formula as T = 2∏√(m/3k). This is the same formula that you found, and it represents the time it takes for the object to complete one full oscillation.

In summary, when an object is attached to multiple springs, the natural frequency and time period of