The discussion revolves around a pendulum clock and how its period affects time measurement. The original poster presents a problem involving the relationship between the period of a pendulum and the time it records over a specified duration.
The discussion includes attempts to clarify the mechanics of the pendulum and its time measurement. Some participants express understanding of the relationship between the number of oscillations and the recorded time, while others seek confirmation of their reasoning regarding the constancy of the time per oscillation.
There is a mention of confusion regarding the mechanism of the pendulum and how it records time, indicating that assumptions about the constancy of the time per oscillation are being examined.
mfb said:Each period of the pendulum, the display of the clock goes forwards by T0.
After N periods, what does the clock show?
Right.DorelXD said:Ohhhhh I get it now. I looked more closely in the mechanism of the pendulum. From what I understood, each time an oscillation is completed the pendulum records a certain time. Let this time be t. This t is constant, and its typical for every pendulum, right ?
In our problem the period , i.e. the time needed for an oscillation to be completed , is modified. But, because our t is a constat, the pendulum will record the same time for each oscilation, even if the number of oscilations increases or decreases.
Correct.In our problem:
In a time D, the pendulum swings : N=\frac{D}{T} times => the pendulum measures the time Nt .
Who is t ? Well we know, from the hypothesis that \frac{D}{T_0}t=D, that is , if the period is T_0 then the time measured by the pendulum is D. Solving for t, we obtain: t = T_0 .
So, Nt = NT_0=\frac{D}{T}T_0. This is the time the pendulum measures.