A. Lynda and Mark decide to have a contest to build a precision timing device. L
ID: 2018614 • Letter: A
Question
A. Lynda and Mark decide to have a contest to build a precision timing device. Lynda chooses a simple pendulum and mark chooses a simple mass-spring system. Each device is to have a period of 2.0(s). The devices are to be constructed in Akron, and you are to select the appropriate values of the parameters to make this happenB. Determine whose device would more precise at the equator. you must calculate the respective periods at the equator and compare them with the values you determined in Akron.
Explanation / Answer
A. We know that angular frequency is
= 2/T
and,
spring = (k/m), so Tspring = 2(m/k)
pendulum = (g/l), so Tpendulum = 2(l/g)
I have no idea what the gravitaional acceleration is in Akron, but find it and choose an l that will give you a period of 2s. For the spring, any value of k or m that will give you a period of 2s 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 Akron 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 Akron, so I cannot calculate the difference in T at the equator, but there wouldn't be a difference in T for the spring.