### 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?

Weβre 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 Ξπ₯ 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 Ξπ₯ two β from its equilibrium
length would be find?

So we want to solve for Ξπ₯
two. And to find it, we can recall
Hookeβs law. Hookeβs 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 itβs in
equilibrium, we can say that the gravitational force is equal in magnitude to the
spring force, negative π times Ξπ₯. 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 itβs the same spring times Ξπ₯
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. Weβre left with a ratio π over π
two is equal to the ratio Ξπ₯ over Ξπ₯ two. Rearranging to solve for Ξπ₯ two,
we see it is equal to Ξπ₯ times the ratio of masses π two over π. Weβve been given Ξπ₯, π, and π
two in our problem statement. So weβre ready to plug in and solve
for Ξπ₯ two. When we enter these values on our
calculator, we find Ξπ₯ two is 7.44 centimetres. Thatβs the displacement from
equilibrium of the spring with the second mass on it.