# SHM(mass suspended vertically on spring)

nrweis

## Homework Statement

An object is hung on the end of a vertical spring and is released from rest with the spring unstressed. If the object falls 3.42 cm before coming to rest, find the period of the resulting oscillatory motion.

## The Attempt at a Solution

This was a homework question given and the answer is in the back of the book and it says
T = .262 seconds. Now i have no clue how they got this because I thought that the period was related to the spring constant and the mass but not to the amplitude so how can we find T if neither k nor m is given?

## Answers and Replies

Staff Emeritus
Gold Member
so how can we find T if neither k nor m is given?

Hi nrweis, welcome to PF!

You can deduce something about them from the restoring force that is required to balance the object's weight.

Homework Helper
Welcome to PF.

You don't necessarily need m and k, because don't you also know that

a(t) = ω2*x(t)

Won't a(t) = 9.8 at x/2 from the dead weight equilibrium point, which is at the peak of its oscillation?

Since ω = 2πf = 2π/T ...

nrweis
Ok I thought about it for a while and thought I was on the right track but still got it wrong. This is what i did:

F=k∆x
mg=k∆x

√(g/∆x)= √(k/m)=ω=2π(1/T)

√(g/∆x)= 2π(1/T)

T=2π√(∆x/g)

and I got like .37 seconds.
Was I on the right track?

nrweis
now I'm all confused. I was thinking that delta x was the amplitude

nrweis
ok I just misunderstood the question. When I divided the 3.42 cm by 2 and reentered it to the formula I got the right answer. Thanks guys.

Homework Helper
now I'm all confused. I was thinking that delta x was the amplitude

Yes. Δx is the amplitude, but the x-position function which expresses x as a function of time is only x/2 distant from the mid point.

Congrats on figuring it out.