Near the top of the Citigroup Centre building in New York City, there is an object with mass of four point zero times ten to the fifth kilograms on springs that have adjustable force constants. Its function is to dampen wind-driven oscillations of the building by oscillating at the same frequency as the building is being driven. The driving force is transferred to the object, which oscillates instead of the entire building. What effective force constant should the springs have to make the object oscillate with a period of 2.00 seconds? What energy is stored in the springs for a 2.00-meter displacement from equilibrium?
In this two-part problem, we’re first asked to solve for a force constant — we’ll call it 𝑘 — that the springs will have when the period — we’ll call it 𝑇 — is 2.00 seconds. We’re then asked what energy is stored in the springs; we’ll call it PE sub 𝑠. We’re then asked what energy is stored in the springs — we’ll call it 𝐸 sub 𝑠 — for a 2.00-meter displacement from equilibrium, which will call 𝑥.
We’re told further than the mass of this object that oscillates is 4.00 times 10 to the fifth kilograms; we’ll refer to it as 𝑚. Given this information, let’s begin on part one, solving for 𝑘, the effective force constant.
As we begin, let’s recall that for a harmonic oscillator, such as the object we have here, there is a relationship between the period of oscillation, the mass of the object, and the spring constant. The relationship or equation that ties those three quantities together says that the period 𝑇 is equal to two 𝜋 times the square root of the mass of the object divided by its spring constant.
Let’s apply this relationship to our scenario. Since we want to solve for 𝑘, the spring constant, let’s rearrange this equation so that it reads 𝑘 equals something else. We can start by squaring both sides of the equation. When we do that, two 𝜋 becomes four 𝜋 squared, and the square root and square term over and divided by 𝑘 cancel one another out.
If we now multiply both sides of our equation by 𝑘 divided by 𝑇 squared, then on the right hand-side of an equation, the 𝑘s cancel out; and on the left- hand side, the 𝑇 squared terms cancel out. We are left with an equation that reads 𝑘 equals four times 𝜋 squared times 𝑚 divided by 𝑇 squared.
𝑚 we know is 4.00 times 10 to the fifth kilograms, and 𝑇 is 2.00 seconds. Let’s enter in these values in our equation to solve for 𝑘. 𝑘 is equal to four 𝜋 squared times 4.00 times 10 to the fifth kilograms divided by 2.00 seconds squared.
When we enter these numbers in our calculator, we find a value for 𝑘, the spring constant, of 3.95 times 10 to the sixth newton-meters. That is the value of the spring constant for the period to be 2.00 seconds.
Now we can move on to part two where we’ll solve for the energy in the springs 𝐸 sub 𝑠 given a displacement of the springs of 2.00 meters. This energy in the spring will come from the compression distance and is a potential energy; that is, an energy that could be released to do useful work.
There is a relationship for the potential energy in a spring in terms of the spring constant and its displacement from equilibrium. The potential energy in a spring is equal to one-half the spring constant, 𝑘, multiplied by its displacement from equilibrium, 𝑥, squared.
Applying this equation to our scenario, we’ve solved for 𝑘, the spring constant, in part one, and 𝑥, the spring’s displacement from equilibrium, is given as 2.00 meters. We can now enter these values in for 𝑘 and 𝑥.
The energy stored in the spring is equal to one-half 3.95 times 10 to the sixth newton-meters multiplied by 2.00 meters, squared The product of these numbers is equal to 7.90 times ten to the sixth joules. This is how much energy is stored in a spring due to a compression of 2.00 meters.