Video: Energy in Simple Harmonic Motion

Video Transcript

A string has a force constant of 53.0 newtons per meter. An object, initially at rest, with a mass of 0.960 kilograms is suspended from it. The object descends, stretching the spring, oscillates, and then comes to rest. How much is the spring stretched when the object has come to rest after oscillating? Calculate the decrease in the gravitational potential energy of the object between its position at the point at which it is attached to the unextended spring and its position at the point at which it comes to rest after oscillating. Calculate the energy stored in the spring by its extension.

In this three-part exercise, we first wanna solve for the displacement of the spring after the mass has been hung on it and comes to rest. Next, we want to solve for the change in gravitational potential energy of the mass between the two points from which it is hung on the unextended spring and the point at which it comes to rest. And finally, we wanna calculate the energy stored in the spring by its extension. We’ll call this PE sub 𝑠.

In this scenario, we start off with a vertically oriented spring at its natural, or equilibrium, length. The spring, we’re told, has spring constant 𝑘 of 53.0 newtons per meter. Then, we take a mass, we’ve called it 𝑚, of value 0.960 kilograms and attach it to the free end of the spring. In response to this, the spring stretches, oscillates, and then comes to rest at a displacement we’ve called Δ𝑥 from its original position.

To solve for Δ𝑥, let’s start by considering the forces that are acting on the mass 𝑚. When the mass has come to rest after oscillating on the end of the spring, we know that the gravitational force acting on the mass is equal in magnitude to the spring force on the mass. So, we can write 𝑚 times 𝑔 is equal to the spring constant 𝑘 times Δ𝑥. Rearranging this expression, we see that Δ𝑥 is equal to 𝑚 times 𝑔 over 𝑘, where 𝑔 we’ll treat as exactly 9.8 meters per second squared.

Plugging in for these three values, when we calculate Δ𝑥, we find that, to three significant figures, it’s 17.8 centimeters. That’s the amount the spring is stretched because of the mass 𝑚. Now, let’s move on to solving for the change in gravitational potential energy of the mass 𝑚 over this displacement.

Recalling that, in general, gravitational potential energy is equal to an object’s mass times the acceleration due to gravity times its height relative to some reference level. Since we’re calculating the change in gravitational potential energy of the mass, we can write that it’s equal to 𝑚 times 𝑔 times Δ𝑥. No negative sign needed.

When we plug in for these three values, being careful to write our displacement Δ𝑥 in units of meters, we find that ΔPE sub 𝑔 is 1.67 joules. That’s the change in the mass’ gravitational potential energy. Finally, we want to solve for the potential energy stored up in the spring as result of this extension.

We recall that in general that potential energy is equal to half a spring‘s constant multiplied by its displacement from equilibrium squared. So, we can write that PE sub 𝑠 is equal to one-half 𝑘 times Δ𝑥 squared, where 𝑘 the spring constant is known and Δ𝑥 we solved for earlier. Substituting in for these values using a distance in meters for our displacement Δ𝑥, we find that PE sub 𝑠 is 0.835 joules. That’s how much potential energy the spring has gained by virtue of being extended.

We may wonder why the spring potential energy is not equal to the change in gravitational potential energy. Recall that, before coming to rest, our mass 𝑚 oscillated up and down on the spring. During that up-and-down motion, it lost energy through friction with the air as well as through the spring. That helps account for the difference between these two values.