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A Second’s pendulum gives correct time at 25C The pendulum shaft is thin and is made up of steel. How many seconds will it lose per day at 35C?? (coeff. of linear expansion of steel = 11×10 -5 /C)

A Second’s pendulum gives correct time at 25C 
The pendulum shaft is thin and is made up of steel.
How many seconds will it lose per day at 35C??
(coeff. of linear expansion of steel = 11×10-5/C)

Grade:11

1 Answers

Arun
25763 Points
2 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

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