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A seconds pendulum clock has a steel wire.The clock Shows correct time at 25 degree centigrade. How much time does the clock loose or gain in one week when the temp is increased to 35 degree centigrade

A seconds pendulum clock has a steel wire.The clock Shows correct time at 25 degree centigrade. How much time does the clock loose or gain in one week when the temp is increased to 35 degree centigrade

Grade:12th pass

1 Answers

Arun
25750 Points
6 years ago
Dear student
 
The period of a pendulum is: 

T = 2*π*sqrt( L/g ) 

With thermal expansion, the period will be longer. Let L = L*(1 + Δ) be the pendulum length at 35C. The longer period is: 

T2 = 2*π*sqrt( L*(1 + Δ) /g ) 

The correct number of periods counted per day is: 

N = 60*60*24 / ( 2*π*sqrt( L/g ) ) = 86400 / ( 2*π*sqrt( L/g ) ) 

The clock would count the same number of periods at 35C, but the time taken would be: 

t2 = N*2*π*sqrt( L*(1 + Δ) /g ) 

t2 = 86400 *2*π*sqrt( L*(1 + Δ) /g ) / ( 2*π*sqrt( L/g ) ) 

A lot of stuff cancels, and: 

t2 = 86400 * sqrt( 1 + Δ ) 

At the end of counting N periods at 35C, the clock has taken an amount of time greater than one day by: 

t2 - 86400 = 86400* ( sqrt(1 + Δ ) - 1) 

For brass, the thermal coefficient of expansion is α = 19*10^-6 C^-1. The length of a pendulum at 35C would be: 

ΔL = L1*α*ΔT = L * 19*10^-6 * 10 = L * 19*10-5 

L2 = L + ΔL = L * ( 1 + α*ΔT ) 

In the above derivation, this was expressed as: 

L = L*( 1 + Δ ) 

Δ = 19*10^-5 

So, the time error after N periods (one day) would be: 

Δt = 86400* ( sqrt(1 + Δ ) - 1) = 86400*( sqrt( 1 + 19*10^-5 ) - 1 ) = 8.2 sec
 
Hence in one week = 7 * 8.2 = 57.4 sec
 
Regards
Arun (askIITians forum expert)

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