• Support PF! Buy your school textbooks, materials and every day products Here!

Length of a pendulum

  • Thread starter cap.r
  • Start date
  • #1
67
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 in to 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,
 

Answers and Replies

  • #2
Spinnor
Gold Member
2,102
315
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

?
 

Related Threads for: Length of a pendulum

  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
2
Replies
46
Views
5K
  • Last Post
Replies
6
Views
667
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
0
Views
3K
Replies
1
Views
3K
Replies
3
Views
15K
Top