The document solves two problems regarding a simple pendulum: 1) Given the frequency and gravity on Earth, it calculates the length of the pendulum string as 0.17m. 2) It then calculates the periods of the pendulum on Earth (0.83s) and Mars (1.34s), explaining that the period is longer on Mars due to its lower gravity of 3.71m/s2 compared to Earth's 9.81m/s2.

1. Physics 101 Learning Object 2: Simple Pendulum
Question:
a) The frequency of a simple pendulum is known to be 1.2 Hz. Given the fact
that gravity is 9.81 m/s2, find the length of the string.
b) Say you took the simple pendulum to Mars, where the gravity is 3.71m/s2.
What would be the period of the pendulum on Mars compared to the period of the
pendulum on Earth? Why is there a difference?
Answer:
a) Step 1:
Since it is a simple pendulum, we know that we use the standard equation for
simple harmonic motion, which is:
ω = √(g/L)
Step 2:
Since we are given the frequency and acceleration due to gravity, we can
use an equation for frequency of oscillation of a pendulum and manipulate it
to solve for L.
f = ω/2π
Now substitute equation from above for omega.
f = (√(g/L))/2π
Then manipulate and to solve for L.
L = g/4π2f2
Step 3:
Now we simply plug in the values of f = 1.2Hz and g = 9.81m/s2 and solve for
the length of the string, L.
L = (9.81m/s2)/4π2(1.2Hz)2
L = (9.81m/s2)/(56.84892135Hz2)
L = 0.1725626409m
L = 0.17m
b) Step 1:
Now this question is asking us about period, but that isn’t a big problem
considering the fact that we know that frequency, f and period, T are inverses
of each other. So, we take the equation of the frequency that we used in part
a) and invert it to get the equation for period, T.

2. f = 1/T
T = 1/f
Now substitute the equation for f that was used in part a) into the equation.
T = 1/((√(g/L))/2π)
Now just rearrange, so it doesn’t look so complicated and messy.
T = 2π(√(L/g)
Step 2:
Now, to find the period of the pendulum on Earth, we simply plug in the
length, L, that we found in part a) which was 0.17m and the acceleration due
to gravity on Earth, gEarth, which is 9.81m/s2.
TEarth = 2π(√(L/gEarth)
TEarth = 2π(√((0.17m)/(9.81m/s2)))
TEarth = 2π(0.1316406315s)
TEarth = 0.8271224817s
TEarth = 0.83s
Step 3:
Now that you’ve found the period of the pendulum on Earth, you need to find
the period of the pendulum on Mars. You use the exact same equation as
above and plug in the value for length, L, from part a) which was 0.17m and
the acceleration due to gravity on Mars, gMars, which is 3.71m/s2.
TMars = 2π(√(L/gMars)
TMars = 2π(√((0.17m)/(3.71m/s2)))
TMars = 2π(0.2140609783s)
TMars = 1.344984794s
TMars = 1.34s
Step 4:
Now just answer the question in words.
The period of the pendulum on Mars is 1.34s compared to the period of
the pendulum on Earth, which is 0.83s. The reason for the pendulum
having a longer period on Mars compared to Earth is because the
acceleration due to gravity on Earth (9.81m/s2) is faster than that on
Mars (3.71m/s2).