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

An object of mass π = 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 π 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?

Weβ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 β we can call Ξπ₯ β of 4.89 centimetres beyond its natural length. What we then want to know is if we attached a different mass β call it π sub two β of 505 grams to the same spring, then what extension of the spring β we can call it Ξπ₯ two β from its equilibrium length would be find?

So we want to solve for Ξπ₯ two. And to find it, we can recall Hookeβs law. Hookeβ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 π is hanging from the spring and itβs in equilibrium, we can say that the gravitational force is equal in magnitude to the spring force, negative π times Ξπ₯. If we were to then write a similar equation, but this time for the second mass π two, it would read π two times π equals negative π the same spring constant because itβs the same spring times Ξπ₯ two.

If we then divide these equations by one another, we see that, on the left-hand side, the factor of π cancels out and, on the right-hand side, negative π the spring constant cancels. Weβre left with a ratio π over π two is equal to the ratio Ξπ₯ over Ξπ₯ two. Rearranging to solve for Ξπ₯ two, we see it is equal to Ξπ₯ times the ratio of masses π two over π. Weβve been given Ξπ₯, π, and π two in our problem statement. So weβre ready to plug in and solve for Ξπ₯ two. When we enter these values on our calculator, we find Ξπ₯ two is 7.44 centimetres. Thatβs the displacement from equilibrium of the spring with the second mass on it.