Help with a simple harmonic motion problem for CB$$$ (in Off-topic)

I've been working on my physics homework all day and have all the questions complete except for one of them, if you can help me out I'll give you 50k cb1 or 2.5k cb2.

Suppose that an object on a vertical spring oscillates up and down at a frequency of 5.1 Hz. By how much would this object, hanging at rest, stretch the spring?

I have similar Physics homework but i'm affraid that you need at leas 1 additional help.

either a mass of the object or something else that you can fiddle it out with.

only thing i can calculate with the 5.1 Hz is the vibrating time of 0.19 seconds

Assuming the spring is on earth, or anywhere else in the universe with a similar gravity, I would say 9.5 mm.

Sorry Bartjan, but [Beshkzu]Dgtwurj_Yxwsi beat you to it via the CB1 post. 2.5k CB2 has been sent to Ex-Spid

It's been far too many years since I've had a physics course (nay, too many since college, even), but I'm curious as to how you figured that out, bart. The relevant formulas escape me.

First, you assume the spring to be "perfect" (i.e. no deformation anomalies, zero mass, frictionless motion, etc).

We use the following notation:

k - Spring constant (unknown)
X - spring elongation "on rest" (solution sought)
dx - momentary spring elongation
dv - momentary mass velocity
g - Earth's gravitational "constant" (known, 9.8 m/sec^2)
m - mass of body (unknown)
f - oscillation frequency (known, 5.1 Hz)
PI - 3.1415... (well known)
T - oscillation period (know, 1/f = 10/51 sec ~- 0.196 sec )

Now, a few words... the oscillation period of a mass on a spring depends ONLY on the weight of the body (m*g) and the spring's strength (k). It doesn't matter how hard it oscillates (1 cm or 1 m), as long as the spring is not deformed (k is constant), the oscillation frequency/period will always be the same. Also, it doesn't matter if we have a gravitation at all, the oscillation period is the same, only the "rest point" or "middle point" is different.

Ok, now, as we have no friction, the energy must conserve.
At the middle point (dx=0) we have no "potential energy", but we have "kinetic energy" (m*dvmax^2).
At the farthest point (each direction), we have dx=max (max potential energy) and dv=0 (no kinetic energy). The potential energy "stored" is equal to the product of distance traveled and force applied ; the force applied went from 0 to k*dxmax linearly, hence we have a potential energy of 0.5*k*dxmax^2.

Anyway, we don't need all of that.
We just have to find out how long does it need to oscillate.
The oscillation can be partitioned in 4 steps, each equal in time (0 -> dxmax -> 0 -> -dxmax -> 0).

At a certain elongation dx, the force pulling the body is k*dx, which translates into an acceleration of k*dx/m (so directly dependant on the elongation).

... Long story short (too bored to type this much and re-calculate each step), the frequency / period of a oscillating mass on a spring is given by the formula:
1/f = T = 2 * PI * sqrt(m/k)

Also, at "rest point", you have m*g = X*k, therefore X = g* (m/k).
You extract m/k as being [T/(2*PI)]^2 ~= 0.974/1000 sec^2, multiply with g (9.8 m/sec^2) and get ~9.54/1000 m.

Hence the ~= 9.54 mm elongation at "rest point" answer.

Or, if you prefer Google's calc: everything before "*g"
and then multiply it with G (~=9.8 m/sec^2) yourself (Google's calc seems to have some issues with the gravitational constant).

Cool; thanks. I will have to ponder this some; as I say, it's been a few years (decades?)

Well, it's been about 12 years ago for me too ;)

oh cmon guys.. i find one place in the world where i dont have to study and here i see it again. Tibiki I hope you are not a teacher...?