# Stuck with a problem on linear thermal expansion of solids

stuck with a problem....on linear thermal expansion of solids

## Homework Statement

I just read a problem 2day...it's on linear thermal expansion of soild...can anyone guide me through....it says..." A sqaure ABCD is made with two different metals...the coefficient of thermal expansion of the metal which made AB & CD is alpha (don't know how to type alpha) & the same of the metal which made BC & AD is beta (??)..Now if the temperature of the system is increased by t. then prove the angle between the diagonals will change by (alpha - beta).t/2 " ?????

can u pls help....

thanks

## The Attempt at a Solution

tiny-tim
Homework Helper
Welcome to PF!

Hi subho123! Welcome to PF!

(have an alpha: α )

The square will become a rectangle, and the angle between the diagonals of a rectangle is twice … ?

thanks tiny-tm for replying so quickly & helping regarding the symbols...well i had an attempt at this...but the value that i m getting...is (α - B).t which is not desired one....let me tell u what i have done...

let the length of the each side of the square is x...then for the increasement of temp by t the side AB & CD will bcom x(1+α.t) & similarly BC & AD will bcom x(1+β.t)..now after temp increase the sqaure will bcome a rectangle ABCD. let the angle DBC is θ. Now, if the diagonals intersect at P, then clearly the angle DPC is 2θ. and the angle DPC has increased by (2θ - Π/2) for the tmp increased.
Now from rectangle ABCD we can say..tanθ = x(1+α.t)/x(1+β.t) = (1+α.t)/(1+β.t)= (1+α.t)(1 +β.t)-1 = (1+α.t)(1 -β.t) = {1+(α-β).t} [as α & β are very small so their higher powers are negligible]

now tan2θ = 2{1+(α-β).t} /1 - {1+(α-β).t}2 = -{1+(α-β).t} /(α-β).t}

now let the change in angle = A = (2θ - Π/2)

tanA = tan((2θ - Π/2)= -tan(Π/2 - 2θ) = -cot2θ = (α-β).t /{1+(α-β).t} = (α-β).t{1-(α-β).t} = (α-β).t i have tried upto this is ...now this isn't the answer...will anybody help me...

thanks

tiny-tim
Homework Helper
Hi subho123!

hmm … i make it the same as you …

tanθ = 1 + (α-β).t …

then using calculus (if you haven't done calculus, never mind, it should give the same answer, so it's a good way of checking) …

sec2θ dθ/dt = (α-β), and sec2θ is approximately 2 (because θ is approximately 45º),

so 2dθ/dt = (α-β),

and so the change in 2θ is approximately (α-β)t

Thanks tiny-tm for the clarification...it seems that the sum has a printing mistake........