# Simple Harmonic Motion & Pendulum Problem

## Homework Statement

What length must the pendulum be changed to in order to show the correct time?
L=0.5m
After 12hours the clock is behind by 30minutes.

w=sqrt(g/L)
w=2πf

## The Attempt at a Solution

I thought if I set the frequency equal to 1Hz and solved for length it would work.

2πf=sqrt(g/L)
4π^2=g/L <-got rid of f since its equal to 1
L=g/(4π^2)
This unfortunately doesn't give the correct answer, I also tried making a ratio involving the time but that failed miserably.

ehild
Homework Helper

## Homework Statement

What length must the pendulum be changed to in order to show the correct time?
L=0.5m
After 12hours the clock is behind by 30minutes.

w=sqrt(g/L)
w=2πf

## The Attempt at a Solution

I thought if I set the frequency equal to 1Hz and solved for length it would work.
Why should be the frequency of the pendulum 1 Hz?

Why should be the frequency of the pendulum 1 Hz?
I thought that if frequency is 1Hz then it would be doing one cycle per second which would result in 1 tick of the clock per second but since its for the entire cycle and each cycle contains 2 ticks? So I should set frequency to 0.5Hz?

ehild
Homework Helper
The swing of the pendulum drives the hands of the clock through some rather complicated mechanical system, including cogwheels. http://visual.merriam-webster.com/s...easure-time/weight-driven-clock-mechanism.php. The time period of the bob is proportional to the period of the hands of the clock, but is not equal to it. In case of a 0.5 m pendulum, the time period is 1.41 s. But the transmission system makes the clock going too slow - by what percent? By what percent have you change the time period of the pendulum, and what does it mean on the length?

The swing of the pendulum drives the hands of the clock through some rather complicated mechanical system, including cogwheels. http://visual.merriam-webster.com/s...easure-time/weight-driven-clock-mechanism.php. The time period of the bob is proportional to the period of the hands of the clock, but is not equal to it. In case of a 0.5 m pendulum, the time period is 1.41 s. But the transmission system makes the clock going too slow - by what percent? By what percent have you change the time period of the pendulum, and what does it mean on the length?

tclock=41 400s
t=43 200s
tclock/t=23/24

The clock is going (1/24)% slower than the actual time, to fix this I should multiply the period by (23/24)?
T=1.41*(23/24)≈1.351

Then w=sqrt(g/L) ->w=2pi/T
(2pi/T)^2=g/L
L=g/(2pi/T)^2=0.453m

Thanks!!!

ehild
Homework Helper
tclock=41 400s
t=43 200s
tclock/t=23/24

The clock is going (1/24)% slower than the actual time, to fix this I should multiply the period by (23/24)?
T=1.41*(23/24)≈1.351

Then w=sqrt(g/L) ->w=2pi/T
(2pi/T)^2=g/L
L=g/(2pi/T)^2=0.453m

Thanks!!!
The method is correct, but you but you made it very complicated, and rounded off too early.
You know that Tcorrect/T = 23/24. Tcorrect/T = sqrt(Lcorrect/0.5). How do you get the correct length from here?

The method is correct, but you but you made it very complicated, and rounded off too early.
You know that Tcorrect/T = 23/24. Tcorrect/T = sqrt(Lcorrect/0.5). How do you get the correct length from here?

Yea that is definitely an easier way to solve for Lcorrect but I'm curious as to how you found this relationship between the period and length.

Nathanael
Homework Helper
Yea that is definitely an easier way to solve for Lcorrect but I'm curious as to how you found this relationship between the period and length.
$\frac{T_a}{T_b}=\frac{2\pi\sqrt{L_a/g}}{2\pi\sqrt{L_b/g}}=\sqrt{\frac{L_a}{L_b}}$

ehild
Homework Helper
Yea that is definitely an easier way to solve for Lcorrect but I'm curious as to how you found this relationship between the period and length.
You know how the formula ##ω=\sqrt{\frac{g}{L}}## was derived and you know the relation between angular frequency and time period: ##\omega=\frac{2\pi}{T}##.
So ##\frac {2\pi}{T}=\sqrt{\frac{g}{L}}##,
isolate T:
##{T}=2\pi\sqrt{\frac{L}{g}}##