A pendulum clock with a brass suspension system is calibrated so that its period

Question

A pendulum clock with a brass suspension
system is calibrated so that its period is 1 s at
18?C.
If the temperature increases to 40 ?C, by
how much does the period change? Answer in
units of s.
I have found the length of the suspension by using the equationt=2L/g and it comes out to be .2482 m Ibelieve. Then I used this equation to find the change in theperiod. 2Lo(1+T)/g andcame up with the answer of 1.000134667 so the change of the periodwould be .000134667 seconds but apparently it is the wrong answercan someone help me with this problem?

Explanation / Answer

Given : A pendulum clock with a brass suspension
system is calibrated so that its period is 1 s at18oC      ( T i ) = 18oC      ( Tf ) = 40oC g = 9.8 m/s2    t = 2 L/g    L = t / 2 * g      Then L = 0.2484 Change in temperature ( T ) = Tf - Ti = 40 -18 =22oC    Coefficient of linear thermal expansion ofbrass ( ) = 19 * 10-6 /oC Time period ( T ) =  2Lo(1+T) /g                            = 2 ( 0.2484 ) ( 1+ 19 * 10-6/oC ( 22oC) / 9.8 m/s2 Time period ( T ) =  2Lo(1+T) /g                            = 2 ( 0.2484 ) ( 1+ 19 * 10-6/oC ( 22oC) / 9.8 m/s2                           = ----- s Solve the above    I hope it helps you          

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