Simple Harmonic Motion & Pendulum Problem

In summary: T1/T2 is the same thing as Tcorrect/T, and L2=0.5m. You get the idea.In summary, to find the correct length of a pendulum in order to show the correct time, you can use the formula Tcorrect/T = sqrt(Lcorrect/0.5), where Tcorrect is the actual time period and T is the given time period. This formula is derived from the relationship between angular frequency and time period, and the formula for angular frequency. This method can be used to calculate the correct length in situations where the clock is behind or ahead by a certain amount of time.
  • #1
Potatochip911
318
3

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.

Homework Equations


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.
 
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  • #2
Potatochip911 said:

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.

Homework Equations


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?
 
  • #3
ehild said:
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?
 
  • #4
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?
 
  • #5
ehild said:
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!
 
  • #6
Potatochip911 said:
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?
 
  • #7
ehild said:
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.
 
  • #8
Potatochip911 said:
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.
[itex]\frac{T_a}{T_b}=\frac{2\pi\sqrt{L_a/g}}{2\pi\sqrt{L_b/g}}=\sqrt{\frac{L_a}{L_b}}[/itex]
 
  • #9
Potatochip911 said:
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}}##
 

Related to Simple Harmonic Motion & Pendulum Problem

1. What is Simple Harmonic Motion (SHM)?

Simple Harmonic Motion (SHM) is a type of periodic motion in which the restoring force is proportional to the displacement from equilibrium and is directed towards the equilibrium point. This results in a repetitive back-and-forth motion around the equilibrium position. Examples of SHM include the motion of a pendulum, a mass on a spring, or a swinging door.

2. How does a pendulum exhibit SHM?

A pendulum exhibits SHM because it follows the basic principles of SHM: a restoring force (gravity) that is proportional to the displacement from equilibrium (height of the pendulum) and is directed towards the equilibrium position (bottom of the swing). As the pendulum swings back and forth, it passes through the equilibrium position and reaches maximum displacement at either end of its swing.

3. What is the relationship between period and length in a simple pendulum?

The period (T) of a simple pendulum is directly proportional to the square root of its length (L), and is independent of the mass and amplitude of the pendulum. In other words, the longer the pendulum, the longer its period. This relationship is described by the equation T = 2π√(L/g), where g is the acceleration due to gravity.

4. How does the amplitude affect the period of a pendulum?

The amplitude (A) of a pendulum, which is the maximum displacement from equilibrium, has no effect on the period. This means that a pendulum with a small amplitude will have the same period as a pendulum with a larger amplitude, as long as the length and gravitational acceleration remain constant.

5. How does the mass of a pendulum affect its period?

The mass (m) of a pendulum has no effect on its period. This is because the restoring force of a pendulum is determined by gravity, which is a constant, and the mass does not affect the gravitational acceleration. Therefore, a pendulum with a heavier mass will have the same period as a pendulum with a lighter mass, as long as the length and gravitational acceleration remain constant.

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