Video: Extension of a Spring Produced by a Particular Mass

An object of mass π‘š = 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?

02:24

Video Transcript

An object of mass π‘š 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 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‘š 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 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 π‘š 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 π‘˜ times Ξ”π‘₯. If we were to then write a similar equation, but this time for the second mass π‘š two, it would read π‘š two times 𝑔 equals negative π‘˜ the same spring constant because it’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 𝑔 cancels out and, on the right-hand side, negative π‘˜ the spring constant cancels. We’re left with a ratio π‘š over π‘š 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‘š two over π‘š. We’ve been given Ξ”π‘₯, π‘š, and π‘š two in our problem statement. So we’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. That’s the displacement from equilibrium of the spring with the second mass on it.

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