Part A Here on Earth, a simple pendulum of length L is observed to have a freque
ID: 1702575 • Letter: P
Question
Part A Here on Earth, a simple pendulum of length L is observed to have a frequency f. If another pendulum on Earth is observed to have a frequency that is 64f, what is its length?64L
8192L
L/4096
L/64
8L
L/8
4096L
L/8192
Part B Here on Earth, a simple pendulum of length L is observed to have a frequency f. On another planet a pendulum of length L is observed to have a frequency of 11f. What is the acceleration due to gravity on this planet? In the answers below, g = 9.8 m/s2.
11g
3.317g
g/242
242g
g/121
g/3.317
121g
g/11
Explanation / Answer
Part-A:
Frequency of a simple pendulum, f = ( 1 / 2 ) ( g / L )
f ( 1 / L )
f^2 1 / L
L ( 1 / f )^2
( L2 / L1 ) = ( f1 / f2 ) ^2
L2 / L = ( f / 64 f ) ^2
L2 = L / 4096
Part - B:
Frequency of a simple pendulum, f = ( 1 / 2 ) ( g / L )
f g
g f^2
( g2 / g1 ) = ( f2 / f1 )^2
( g2 / g1 ) = ( 11f / f )^2
g2 = 121 g