2 quick pendulum questions

1. Jan 16, 2004

nation_unknown

Hey. I would like to thank everyone in advance for reading my questions and for helping me. I am really greatful for it. I have probably put 3 or 4 hours into these questions but just can not seem to be able to break them down and figure out how to come up with a solution.

Question 1: A pendulum which is 2.00m long makes 84.5 compleate oscillations in 4.00 min. What is the acceleration of gravity at this spot? Show all work and formulas.

Question 2: A grandfather clock uses a large mass suspended at the end of a metal rod as a timing mechanism. The back and forth movement of the mass provides the ticks and tocks for this clock.
An interesting characteristic of this type of clock is that it is sensitive to temperature. Imagine that three identical grandfather clocks are set to the same time and then allowed to run for one week. The clock on the left ran in a room that was hot. The clock on the right wan in a room that was cold. The clock in the center ran in a room that was room temprature.
Sketch the approximate times that the clocks in the hot and cold rooms would show. Explain your answer.

(clock at room temprature points to exactly 9 o'clock).

Thank-you so much for your time and help. any help givin would give me a great boost here :).

2. Jan 16, 2004

HallsofIvy

Staff Emeritus
You've put in 3 or 4 hours into these questions? Don't you have some work that you can show us? What did you do all those hours?

1) The clock has done "84.5 compleate oscillations in 4.00 minutes" so its period is 4/84.5= 0.047 minutes per oscillation= 2.84 oscillations per second. Do you know a formula for the period of a pendulum?

2) You know, I presume that the formula for the period of a pendulum involves the length of the pendulum (that's why, in problem 1, they have to tell you that the length is 2.00 meters). What normally happens to the length of something (metal,say) when you heat it? When you cool it?

3. Jan 19, 2004

nation_unknown

Sorry for not getting back to you sooner about this, it has been a very busy weekend. For number one i figured that the answer to the problem would have something to do with the formula g=GM/R^2 (i couldnt get the latex thing to work right)where g=gravity, G=6.67x10^-11, M=mass of the pendulum, and r=the radius of the pendulum. From there i tried to figure out what the mass and the radius of the pendulum was based on the information givin however i couldnot come up with any ways to find these. I then tried to work with centripetal acceleration through the equation:
ac=4(pi)^2r/T^2 however even with this i could only get far enough to find everything except the radius (r).

As for your question about me knowing the formula for the period of a pendulum, from what I understand it is T=2(pi)squareroot(l/g)however i can not seem to find g in this question [b(]

The second question (2) i understand a little better now due to your explanation. I was totally stumped before and tried things such as g=GM/R^2 however there was no way to solve it like that. I understand that heat speeds up (expands) the molecules in metal therefore making the length of the pendulum longer and that cold tempratures makes metal contract therfore making the pendulum shorter. Based on T=2(pi)squareroot(l/g) then the clock in a hot room would be a little behind the time of the clock in room temperature because of the extended period and the clock in the cold room would be a little ahead of the clock at room tempertaure because the pendulum would be shorter making for a shorter time period, making isolations shorter. Thank-you for your help on this question but i still am having quite a bit of trouble understanding the first one.

4. Jan 19, 2004

Warr

if you have this equation

$$T=2\pi\sqrt{\frac{L}{g}}$$

then you can rearrange for g, and substitute in L and T

also you will want to convert your period into seconds/oscillation, since acceleration (g) is expressed in m/s rather than m/min. Therefore the period is (0.047*60) seconds/oscillation = 2.82 seconds/oscillation

$$g=\frac{L}{(\frac{T^2}{4\pi^2})}$$

Last edited: Jan 19, 2004
5. Jan 19, 2004

nation_unknown

Thank you very much for your time. I thought it might be something simple like that. I can't believe I forgot what g was.