Calculate Fraction to Change Length of Pendulum for True Run

  • Thread starter Thread starter cap.r
  • Start date Start date
  • Tags Tags
    Length Pendulum
Click For Summary
SUMMARY

The discussion centers on calculating the necessary fraction to change the length of a pendulum to correct for a time loss of one minute every hour. The key equation used is T = 2π/√g * √l, where T represents the period and l the length of the pendulum. The correct approach to adjust the period is T = t(1 - 1/60), indicating that the period must be shortened by a fraction of 1/60 to achieve the desired frequency of 60 swings per minute. This adjustment ensures the pendulum runs true despite the initial time loss.

PREREQUISITES
  • Understanding of pendulum motion and period calculations
  • Familiarity with the equation T = 2π/√g * √l
  • Basic knowledge of frequency and its relationship to period
  • Concept of fractional changes in physical measurements
NEXT STEPS
  • Study the derivation of the pendulum period formula T = 2π/√g * √l
  • Explore the concept of frequency and its calculation in oscillatory systems
  • Learn about the effects of length changes on pendulum motion
  • Investigate practical applications of pendulum timing corrections in clocks
USEFUL FOR

Students studying physics, particularly those focused on mechanics and oscillatory motion, as well as educators seeking to explain pendulum behavior and timekeeping principles.

cap.r
Messages
64
Reaction score
0

Homework Statement


by what fraction must you change the length of the pendulum to make it run true if it loses one minute every hour

Homework Equations



t=2pi/sqrt(g) * sqrt(l)

The Attempt at a Solution


I know that I am losing 1 second every 60 seconds so I need to shorten the period by a certain amount so that it swings 60 times every 60 second instead of the 59 times that it currently swings.

so I want to say T=t-1/60 and then plug that into find L (T and L are final while t and l are initial).

but the correct answer is T=t(1-1/60), I don't know why this is or how we get it. I can do the rest of the problem if I can understand this one thing.

thanks,
 
Physics news on Phys.org
Does this work?

The frequency f(L) is 1/T(L) = [1/2pi]*[g/L]^.5 = 3540/3600 we get 3540 cycles per 3600 seconds with length L. But we want,

f(L-L/a) = [1/2pi]*[g/L-L/a]^.5 = 3600/3600

?
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 32 ·
2
Replies
32
Views
3K
  • · Replies 7 ·
Replies
7
Views
4K
Replies
23
Views
2K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 20 ·
Replies
20
Views
2K
Replies
1
Views
1K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 26 ·
Replies
26
Views
4K
Replies
2
Views
1K