# Video: Extension of a Spring Produced by a Particular Mass

An object of mass 𝑚 = 332 g is suspended from a vertically hanging spring, causing the spring to extend by 4.89 cm. What extension of the spring would be produced by suspending an object with a mass of 505 g from it?

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### Video Transcript

An object of mass 𝑚 equals 332 grams is suspended from a vertically hanging spring, causing the spring to extend by 4.89 centimetres. What extension of the spring would be produced by suspending an object with a mass of 505 grams from it?

We’re told in this example that if we start out with a spring hanging under its own weight and then attach a mass of value 332 grams to the end of the spring, the spring will extend a distance — we can call Δ𝑥 — of 4.89 centimetres beyond its natural length. What we then want to know is if we attached a different mass — call it 𝑚 sub two — of 505 grams to the same spring, then what extension of the spring — we can call it Δ𝑥 two — from its equilibrium length would be find?

So we want to solve for Δ𝑥 two. And to find it, we can recall Hooke’s law. Hooke’s law tells us that the restoring force exerted by a spring is equal to negative its spring constant times its displacement from equilibrium. If we consider the forces that are acting on the spring when the mass 𝑚 is hanging from the spring and it’s in equilibrium, we can say that the gravitational force is equal in magnitude to the spring force, negative 𝑘 times Δ𝑥. If we were to then write a similar equation, but this time for the second mass 𝑚 two, it would read 𝑚 two times 𝑔 equals negative 𝑘 the same spring constant because it’s the same spring times Δ𝑥 two.

If we then divide these equations by one another, we see that, on the left-hand side, the factor of 𝑔 cancels out and, on the right-hand side, negative 𝑘 the spring constant cancels. We’re left with a ratio 𝑚 over 𝑚 two is equal to the ratio Δ𝑥 over Δ𝑥 two. Rearranging to solve for Δ𝑥 two, we see it is equal to Δ𝑥 times the ratio of masses 𝑚 two over 𝑚. We’ve been given Δ𝑥, 𝑚, and 𝑚 two in our problem statement. So we’re ready to plug in and solve for Δ𝑥 two. When we enter these values on our calculator, we find Δ𝑥 two is 7.44 centimetres. That’s the displacement from equilibrium of the spring with the second mass on it.