Bart Simpson and Lisa Simpson decide to have a contest to build a precision timi

Question

Bart Simpson and Lisa Simpson decide to have a contest to build a precision timing device. Bart chooses a simple pendulum, and Lisa chooses a simple mass-spring system. Each device is to have a period of 3.0 (s). The devices are to be constructed in Springfield, and you are to select the appropriate values of the parameters to make this happen.

b) Determine whose device would be more precise at the equator. You must calculate the respective periods at the equator, and compare them with the values youd etermined in Springfield.

Gravity at the equator= 9.789m/s^2

Explanation / Answer

A. We know that angular frequency is ? = 2p/T and, ?spring = v(k/m), so Tspring = 2pv(m/k) ?pendulum = v(g/l), so Tpendulum = 2pv(l/g) The problem with this question is that they never explicitly say which Springfield the Simpsons are based in on the show, so unless you were given the acceleration due to gravity in the question, you could not actually be expected to solve this problem. Once you knew the gravitational acceleration in Springfield, use it in the above equation and fit l to give you T = 3s. For the spring, any value of k or m that will give you a period of 3s is fine. B. The difference between the two methods of keeping time is that the pendulum depends upon the LOCAL gravitational acceleration, whereas the spring is independent of any outside influences (it only depends upon its OWN mass and its OWN spring constant), so building a pendulum timer in Springfield and then taking it to the equater is not as accurate as building a spring timer and taking it to the equater, because the pendulum's period will change since the gravitational acceleration would have changed (you can bet than any latitutde change such as that would change the gravitiational acceleration, becaus unless the elevation changes just enough to cancel out the change in latitude, it will always change the value of g). So, the more precise method would be the spring. Once again, I do not know the value of g in Springfield, so I cannot calculate the difference in T at the equator, but there wouldn't be a difference in T for the spring.

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