# Linear expansion of a pendulum

1. Jan 26, 2014

### castrodisastro

1. The problem statement, all variables and given/known data
A clock based on a simple pendulum is situated outdoors in Anchorage, Alaska. The pendulum consists of a mass of 1.00kg that is hanging from a thin brass rod that is 2.000m long. The clock is calibrated perfectly during a summer day with an average temperature of 19.5°C. During the winter, the average temperature over one 24-h period is −25.5°C. Find the time elapsed for that period according to the simple pendulum clock.

Tinitial = 19.5 °C = 292.5 K
Tfinal = -25.5 °C = 247.5 K
ΔT = -45 K
mp = 1.00 kg
Lrod = 2.00 m
αbrass=19*10-6/°C

2. Relevant equations
Ltotal=L+α(ΔT)L
T=2$\pi\sqrt{}(L/G)$

3. The attempt at a solution
I calculate the change in length due to the linear expansion of the brass rod

Ltotal=(2.00 m)+(19*10-6)(45 K)(2.00 m)
Ltotal=(2.00 m)+(-1.71*10-3)
Ltotal=(1.9983 m)

The I calculated the length of the period.

T=2pi(1.99/(9.81m/s2))$\frac{1}{2}$
T=2pi(0.451 s)
T=2.8358 s

I submitted this answer to my online homework but it was incorrect. I think it has something to do with how the question includes "...winter, the average temperature over one 24 -h period is -25.5 °C. Find the time elapsed for that period according..." but I don't understand if it is asking for something more.

Any Help is appreciated.

2. Jan 26, 2014

### vela

Staff Emeritus
You would be claiming, if your answer were correct, that over a 24-hour period, only 2.8358 second elapses on the clock. Doesn't sound right, does it?

3. Jan 26, 2014

### castrodisastro

So if the rod shortens, the period of the pendulum would be shorter, the time elapsed on the clock during a 24 hour period would more than 24 hours.

If the period was originally 2.8370, it is now 2.8358

If a 24 hour period is 86,400 seconds, one period of the pendulum is 2.8370 seconds, it would take

(86,400 s)/(2.8370 s/period) = 3,0454.705 periods originally.

Would I then multiply 3,0454.705 periods by the new length of the period 2.8358 seconds to see how many seconds would elapse on the clock after the same number of periods required for 24 hours before the rod contracted.

(3,0454.705 periods)(2.8358 s/period) = 86,363.45244 s

(86,363.45244 seconds)/(3600s/h) = 23.9898 hours

So does this mean that after 24 hours, the clock would read 23 hours 59 hours 23.5 seconds.?

4. Jan 26, 2014

### vela

Staff Emeritus
You kinda have the right idea, but you're holding the wrong quantity constant. It's not the number of periods that stays the same; it's the actual time elapsed that remains at 24 hours for both cases. You want to calculate the number of shortened-rod periods that fit into that time interval and then figure out what reading that corresponds to on the clock.

5. Jan 26, 2014

### castrodisastro

If the period of the shortened rod is 2.8358 seconds, then during a 24 hour day, the pendulum experiences 3,0467.593 periods.

3,0467.593 periods is 100.0423% of the original number of periods. Would this mean that the clock would read

24 h + (24 h*100.0423%) = 24 hours, 0 minutes, and 36.56 seconds

That makes sense, hopefully that is correct.

6. Jan 26, 2014

### castrodisastro

Thanks, I just submitted that answer to my online homework and it was correct. I appreciate the help

7. Jan 26, 2014

### vela

Staff Emeritus
You're welcome. Glad you figured it out.