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Problem wit h a pendulum

  • Thread starter DorelXD
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  • #1
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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
 

Answers and Replies

  • #2
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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?
 
  • #3
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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?
Ohhhhh I get it now. I looked more closely in the mechanism of the pendulum. From what I understood, each time an oscilation 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 oscilation 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 belive 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.
 
  • #4
34,070
9,957
Ohhhhh I get it now. I looked more closely in the mechanism of the pendulum. From what I understood, each time an oscilation 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 oscilation 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.
 
  • #5
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Thank you very much!!!!
 

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