Physics Challenge Problem-Good problem help?

Question

As the captain of the scientific team sent to Planet Physics, one of your tasks is to measure gravity. You have a long, thin wire labeled 1.15 g/m and a 1.22 kg weight. You have your accurate space cadet chronometer but, unfortunately, you seem to have forgotten a meter stick. Undeterred, you first find the midpoint of the wire by folding it in half. You then attach one end of the wire to the wall of your laboratory, stretch it horizontally to pass over a pulley at the midpoint of the wire, then tie the 1.22 kg weight to the end hanging over the pulley. By vibrating the wire, and measuring time with your chronometer, you find that the wire's second harmonic frequency is 200 Hz . Next, with the 1.22 kg weight still tied to one end of the wire, you attach the other end to the ceiling to make a pendulum. You find that the pendulum requires 313 s to complete 200 oscillations. Pulling out your trusty calculator, you get to work.

What value of gravity will you report back to headquarter

Answer 1

Hi there! It seems obvious from the wording of the problem that you are expected to find the length of the wire from the second harmonic of the wire. These two equations would probably be useful:
> f = n/2L*sqrt(T/mu) where f is the nth harmonic frequency (n is 2 in this case), L is the length of the wire, T is the tension in the wire, and mu is the linear density of the wire; so, f = 200, n = 2, L is unknown, T is unknown, and mu is 1.15 g/m (or 1.15 * 10^-3 kg/m)
> t = 2*pi*sqrt(L/g), where t is the period of oscillation of a pendulum, L is the length of the wire, and g is the gravitational constant we want to find; so, t = 313/200 s (the time to complete one oscillation), L is again unknown, and g is what we need to find.
In attempting to solve the first equation for L, we run into the problem of not knowing the value of T, the tension in the wire. We should be able to calculate this based on the mass of the hanging weight in terms of g. A straightforward T = m*g should work here.
Try to combine the equations into one equation using algebra and solve for g. You may need to square both sides of an equation a couple times.
I'm not sure if L in both equations will be the same, however, since only half of the wire is hanging from the pulley in the first part; so, in the first equation L may be only half the length of the wire. I calculated g for both cases and both of the answers seemed plausible (answers between 1 and 10, not, like, 8.34 * 10^-9 or 1.22 * 10^12 or some other bizarre answer).
Hope that helps you in the right direction!
:)
Davy

Answer 2

Presuming you mean g, as in weight = mg for the "value of gravity, why would I do all that pulley and harmonics stuff?

Why not just use S = 1/2 gt^2; so that g = sqrt(2S)/t. where S is your known height (you had to pass a physical to go on the space flight; so you would know how tall you were), t = the time to drop any object (including that 1.22 kg weight if you wish) from S, as measured by your very accurate Rolex.

I'd do this experiment about 30 times or so, calculate the average g and standard deviation for the population of data, and submit that back to headquarters.

Answer 3

Presuming you imply g, as in weight = mg for the "significance of gravity, why could I do all that pulley and harmonics stuff? Why no longer simply use S = a million/two gt^two; in order that g = sqrt(2S)/t. wherein S is your identified top (you needed to move a bodily to head at the house flight; so that you could know the way tall you had been), t = the time to drop any item (adding that a million.22 kg weight when you want) from S, as measured by means of your very correct Rolex. I'd do that test approximately 30 occasions or so, calculate the ordinary g and normal deviation for the populace of knowledge, and publish that again to headquarters.

Answer 4

I get 5.52m/s^2

You have two unknowns, and need two equations. These are:
1) f=n/(2L)(F/u)^.5 --> fundamental frequency equation for vibration
2) f=1/(2*pi)(a/L)^.5 -->pendulum frequency equation

For the first equation:
f=fundamental frequency (stated as 200Hz in the question)
n=number of harmonic (in this case the second, n=2)
L=length of the string being vibrated (in this case L=L/2 as only one half the length is vibrating)
F=tension in the string = ma (in this case, I neglect the weight of the string since 1.22kg>>.00115kg, so ma=1.22a)
a=acceleration due to the gravity on this planet
u=mass of the string per unit length (.00115 kg/m)

f=n/(2L)(F/u)^.5
200=2/(2L/2)(1.22a/.00115)^.5
solve to get L^2=.106a

For the second equation:
f=pendulum frequency (200/313 for this question)
L=length of pendulum (now L, it is no longer folded in half)
a=gravity on this planet

f=1/(2*pi)(a/L)^.5
200/313=1/(2*3.14)(a/L)^.5
solve to get L=a/16.12

Sub the result from 2 into result from 1 to get:
L^2=(.106a)^2=a/16.12
.011236a^2-.0620a=0
a(.011236a-.0620)=0

roots are:
a=0
a=.0620/.011236= 5.52m/s^2

The formulas and methodology are good, not sure on any math errors in between there and the answer. Work it out.

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