How does the period of a pendulum affect its time measurement?

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Homework Statement



A pendulum clock measures the time exactly if its period is [itex]T_0[/itex]. What time does the pendulum record in a time [itex]D[/itex] , if its period becomes [itex]T[/itex] ?

Homework Equations



I know that the number of oscilations of the pendulum in the time D is : N=D/T

The Attempt at a Solution



Well I don't know how to use the informations that the probelm gives me.
P.S. : SORRY FOR THE SPELLING IN THE TITLE
 
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Each period of the pendulum, the display of the clock goes forwards by T0.
After N periods, what does the clock show?
 
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mfb said:
Each period of the pendulum, the display of the clock goes forwards by T0.
After N periods, what does the clock show?

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 [itex]t[/itex]. This [itex]t[/itex] 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 [itex]t[/itex] is a constat, the pendulum will record the same time for each oscilation, even if the number of oscilations increases or decreases.

In our problem:

In a time D, the pendulum swings : [itex]N=\frac{D}{T}[/itex] times => the pendulum measures the time [itex]Nt[/itex] .

Who is [itex]t[/itex] ? Well we know, from the hypothesis that [itex]\frac{D}{T_0}t=D[/itex], that is , if the period is [itex]T_0[/itex] then the time measured by the pendulum is D. Solving for t, we obtain: [itex]t = T_0[/itex] .

So, [itex]Nt = NT_0=\frac{D}{T}T_0[/itex]. This is the time the pendulum measures.

Please, help me, and tell me if my judgement is correct. I believe that what confused me before was that I wasn't fully aware that the mechanism of a pendulum allows it to record the same amount of time, and that this time ( [itex]t[/itex] ) doesn't depend on the number of oscilations.
 
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 [itex]t[/itex]. This [itex]t[/itex] 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 [itex]t[/itex] is a constat, the pendulum will record the same time for each oscilation, even if the number of oscilations increases or decreases.
Right.

In our problem:

In a time D, the pendulum swings : [itex]N=\frac{D}{T}[/itex] times => the pendulum measures the time [itex]Nt[/itex] .

Who is [itex]t[/itex] ? Well we know, from the hypothesis that [itex]\frac{D}{T_0}t=D[/itex], that is , if the period is [itex]T_0[/itex] then the time measured by the pendulum is D. Solving for t, we obtain: [itex]t = T_0[/itex] .

So, [itex]Nt = NT_0=\frac{D}{T}T_0[/itex]. This is the time the pendulum measures.
Correct.
 
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Thank you very much!
 

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