Thermal Expansion - Pendulum of a Grandfather Clock

[dmullin4]

Thermal Expansion -- Pendulum of a Grandfather Clock

Homework Statement

The pendulum in a grandfather clock is made of brass (coefficient of linear expansion = 19 x 10^-6) and keeps perfect time at 17 degrees Celsius. How much time is gained or lost in a year if the clock is kept at 28 degrees Celsius? (Assume the frequency dependence on length for a simply pendulum applies.)

Homework Equations

T = 2pi * sqrt(L/g)
delta L = alpha*L(naught)*delta T
L = L(naught)*(1 + alpha*delta T)

The Attempt at a Solution

I really don't know where to even start with this. Once given a push in the right direction, I should be ok. But I just don't get the relationship.

 

[Hootenanny]

Welcome to Physics Forums.

HINT: What is the length of the pendulum if it keeps perfect time?

 

[dmullin4]

I'm not sure. Wouldn't it depend on the gravity at a certain location? The instructor posted a worksheet that compares the T = 2pi * sqrt(L/g) initial to the final. But the comparisons he made are foreign to me. He has the fractional loss given by delta T/T(naught). Then he does a lot of substitutions and simplifications that I don't completely follow.

 

[Hootenanny]

I'm not sure. Wouldn't it depend on the gravity at a certain location? The instructor posted a worksheet that compares the T = 2pi * sqrt(L/g) initial to the final. But the comparisons he made are foreign to me. He has the fractional loss given by delta T/T(naught). Then he does a lot of substitutions and simplifications that I don't completely follow.

Okay, let's start by finding the ratio T/T₀ then.

 

[dmullin4]

Ok, so I have T₀ = 2pi * sqrt(L₀/g) at 17^o C and T = 2pi * sqrt(L/g) at 28^o Celsius. T is greater than T₀ because L is greater than L₀. I understand this part. But then they have the clock is losing time by an amount of delta T = T-T₀. Where does that come from?

 

[Hootenanny]

Ok, so I have T₀ = 2pi * sqrt(L₀/g) at 17^o C and T = 2pi * sqrt(L/g) at 28^o Celsius. T is greater than T₀ because L is greater than L₀. I understand this part. But then they have the clock is losing time by an amount of delta T = T-T₀. Where does that come from?

Good. Now, the next step is to find the fractional change in period, as your instructor did. So can you write down

$$\frac{T}{T_0} = \ldots$$

 

[dmullin4]

T/T₀ = 2pi * sqrt(L/g) / 2pi * sqrt(L₀/g)

= sqrt(L/g) / sqrt(L₀/g)

= sqrt(L) / sqrt(L₀)

= ?

Now I'm lost again.

 

[rl.bhat]

T/T₀ = 2pi * sqrt(L/g) / 2pi * sqrt(L₀/g)

= sqrt(L/g) / sqrt(L₀/g)

= sqrt(L) / sqrt(L₀)

= ?

Now I'm lost again.

$$\frac{T}{To}$$ = sqrt( Lo + Lo*α*t)/sqrt(Lo)
Length Lo of the grandfather clock is 1 m and its period is 2 s.
Since $$\alpha$$ is very small, by using binomial expansion, the equation can be rewritten as

$$\frac{T}{To}$$ = 1 + 1/2*α*t, where t is the temperature difference and α is the coefficient of linear expansion.

$$\frac{T}{To}$$ - 1 = 1/2*α*t

$$\frac{T - To}{To}$$ = 1/2*α*t

(T - To)/To = 1/2*α*t = the chage in then period per second.
(T - To) is the change in one oscillation, i.e. in 2 seconds.
( T - To )/To* number of seconds in one year gives you the required result.

 

[dmullin4]

$$\frac{T}{To}$$ = sqrt( Lo + Lo*α*t)/sqrt(Lo)
Length Lo of the grandfather clock is 1 m and its period is 2 s.

Where do you get the information that T = sqrt(L₀ + L₀*a*t) ?

 

[Hootenanny]

Where do you get the information that T = sqrt(L₀ + L₀*a*t) ?

Okay, I'll walk through it slowly. We start from your first line,

T/T0 = 2pi * sqrt(L/g) / 2pi * sqrt(L0/g)

$$\frac{T}{T_0} = \frac{2\pi\sqrt{L/g}}{2\pi\sqrt{L_0/g}}$$

Notice that the factors of 2 pi cancel and we can combine the square root.

$$\frac{T}{T_0} = \sqrt{\frac{L}{g}\frac{g}{L_0}}$$

The factors of g cancel yielding

$$\frac{T}{T_0} = \sqrt{\frac{L}{L_0}}$$

You with me so far?

The next step is to insert the expression for L in terms of L₀, T and a into the expression. Do you follow?

 

[dmullin4]

Yes, I got that far. So now we use L = L₀(1 + alpha * delta T) ?


sqrt(L₀(1 + alpha * delta T)/L₀) = sqrt(1 + alpha * delta T)

alpha = 19 x 10^-6
delta T = 28 - 17 = 11

sqrt(1 + (19 x 10^-6 * 11)) = sqrt(1 + 2.09 x 10^-4)

sqrt(1.000209) = 1.0001044...

So is this number right? If so, now what?

 

[Hootenanny]

Try re-reading rl.bhat's post, paying particular attention to the binomial expansion part.

 

[dmullin4]

$$\frac{T}{To}$$ = sqrt( Lo + Lo*α*t)/sqrt(Lo)
Length Lo of the grandfather clock is 1 m and its period is 2 s.
Since $$\alpha$$ is very small, by using binomial expansion, the equation can be rewritten as

$$\frac{T}{To}$$ = 1 + 1/2*α*t, where t is the temperature difference and α is the coefficient of linear expansion.

$$\frac{T}{To}$$ - 1 = 1/2*α*t

$$\frac{T - To}{To}$$ = 1/2*α*t

(T - To)/To = 1/2*α*t = the chage in then period per second.
(T - To) is the change in one oscillation, i.e. in 2 seconds.
( T - To )/To* number of seconds in one year gives you the required result.

OK, I got the part where sqrt(L₀(1 + alpha * delta T)/L₀) would equal sqrt((L₀ + L₀* alpha * delta T)/T₀).

What I don't understand is where the 1/2 comes from. Does it have something to do with the original length being 1m and the original period equalling 2s?

 

[rl.bhat]

OK, I got the part where sqrt(L₀(1 + alpha * delta T)/L₀) would equal sqrt((L₀ + L₀* alpha * delta T)/T₀).

What I don't understand is where the 1/2 comes from. Does it have something to do with the original length being 1m and the original period equalling 2s?

If x is very small ( 1 + x )^1/2 can be written as 1 + 1/2*x, by neglecting the higher powers of x in the binomial expansion.

 

[dmullin4]

I THINK I GOT IT!

Ok,
Original Period = 2pi*sqrt(L0/g) at 17 Celsius
New Period = 2pi*sqrt(L/g) at 28 Celsius

delta T/T0 = T-T0/T0 = (T/T0)-1

[2pi*sqrt(L0/g) / 2pi*sqrt(L/g)] -1 = sqrt(L/L0) -1 = sqrt(L0(1+alpha*delta temperature)/L0) -1

sqrt(1 + alpha*delta temperature) - 1 = sqrt(1 + 19*10^-6*11) - 1 = 1.044*10^-4

1.044*10^-4 * 31536000s = 3295.339s = 54.9 minutes!

My only question now is: Where did I pull "delta T/T0" from? I saw it on another website and used it to get the right answer, but I don't understand how it fits. I know that since its getting heated, the pendulum will expand, making L > L0 and thus T > T0. Am I missing something elementary? I feel like I'm looking over the obvious.

Thanks for everyone help.

 

[rl.bhat]

sqrt(1 + 19*10^-6*11) - 1 = 1.044*10^-4

Check this calculation.

 

[dmullin4]

sqrt(1 + 19*10^-6*11) - 1 = 1.044*10^-4

Check this calculation.

That caculation checks out. Let me try to type it better.

{sqrt[1+(19*10^-6)*11]} - 1 = sqrt(1.000209) - 1 = 0.0001044 = 1.044*10^-4

 

[rl.bhat]

That caculation checks out. Let me try to type it better.

{sqrt[1+(19*10^-6)*11]} - 1 = sqrt(1.000209) - 1 = 0.0001044 = 1.044*10^-4

OK. That is correct.