The period of a simple pendulum is directly proportional to the
square root of its length. If a pendulum has a length of 6 feet and a
period of 2 seconds, to what length should it be shortened to achieve a 1 second period?
2006-10-26 03:56:00 · 6 answers · asked by Dennis B 1 in Science & Mathematics ➔ Mathematics
period= constant*sqrt(length)
first, find constant, plug in values that you know:
2=constant*sqrt(6)
so constant=2/sqrt(6)
so now you have equation
period= 2/sqrt(6)*sqrt(length)
plug in 1 for period to find the length needed
1=2/sqrt(6)*sqrt(length)
simplify further:
sqrt(6)/2=sqrt(length)
so length has to be equal to 6/4 = 1.5 ft
Hope that helps!
2006-10-26 04:01:42 · answer #1 · answered by D 3 · 1⤊ 0⤋
T=k*sqrt(L)
L=6, T=2, so, 2=k*sqrt(6), so, k=2/sqrt(6)
for, T=1, k=2/sqrt(6), so, sqrt(L)= sqrt(6) / 2
so, L = 6/4=3/2
right? as long as k is constant
2006-10-26 11:02:07 · answer #2 · answered by tsunamijon 4 · 0⤊ 0⤋
[T/root(l)]first = [T/root(l)]later
2 s / V(6 ft) = 1 s / V(x ft) Square left and right
4 / 6 = 1 /x. So x = 1.5
The new length should be 1.5 feet.
Th
2006-10-26 11:04:02 · answer #3 · answered by Thermo 6 · 0⤊ 0⤋
proportion is 2 seconds:square root of 6 feet
divide by two: 1: the sqaure root of six divided by two
the sqaure root of six divided by two equals 1.225
oops sorry my answers wrong. the one above mine is right :)
2006-10-26 11:01:50 · answer #4 · answered by Faz 4 · 0⤊ 0⤋
T = 2*Pi Sqrt(L/g)
= 2 * 3.14159 * Sqrt(6/32)
= 2.72 sec
2006-10-26 11:02:07 · answer #5 · answered by The Potter Boy 3 · 0⤊ 1⤋
2.4494 feet
2006-10-26 11:02:49 · answer #6 · answered by Ashish Samadhia 3 · 0⤊ 1⤋