The Simple Pendulum
Summary
TLDRThis educational video script explains the concept of a simple pendulum, detailing its period and frequency. It emphasizes that the period is the time for one complete swing, while frequency is the number of swings per second. The formula for calculating the period involves the pendulum's length and gravitational acceleration, with the mass of the bob being irrelevant. Examples are provided to illustrate these concepts, including calculating the period and frequency on Earth and the Moon, determining the length of a pendulum given its period, and estimating gravitational acceleration on an unknown planet using a pendulum.
Takeaways
- 🕰 The period (T) of a simple pendulum is the time it takes to complete one full swing from point A to C and back to A.
- 🔄 Frequency is the reciprocal of the period and represents the number of complete swings or cycles per second, measured in Hertz (Hz).
- 📏 The period of a pendulum is determined by its length (L) and the gravitational acceleration (g), with the formula T = 2π√(L/g).
- 🌐 The period of a simple pendulum is independent of the mass of the pendulum bob, as mass is not part of the period calculation equation.
- 🌍 The gravitational acceleration on Earth is approximately 9.8 m/s², and this value is used in the period calculation for pendulums on Earth.
- 🌕 The gravitational acceleration on the Moon is about 1.6 m/s², which is less than Earth's, leading to a longer period for the same pendulum length.
- 📉 As the length of the pendulum string (L) increases, the period (T) increases, and the frequency decreases, showing an inverse relationship.
- 📈 Conversely, as the gravitational acceleration (g) increases, the period (T) decreases, and the frequency increases, also showing an inverse relationship.
- 🔍 To find the gravitational acceleration of an unknown planet, you can use a simple pendulum by knowing its length and the time for a certain number of swings.
- 🕰️ The period of a grandfather clock's pendulum, which has a one-second interval between ticks and tocks, is actually two seconds as it represents a complete swing.
Q & A
What is a simple pendulum and how is it represented?
-A simple pendulum is a weight suspended from a fixed point so that it can swing freely back and forth under the influence of gravity. In the script, it is represented by a vertical line with a pendulum at an angle, marked by points A, B, and C.
What is the significance of a complete swing in a pendulum's motion?
-A complete swing in a pendulum's motion is significant because it allows for the determination of the pendulum's period and frequency, which are essential for understanding its oscillatory behavior.
How is the period of a simple pendulum defined and measured?
-The period of a simple pendulum, represented by the symbol T, is defined as the time it takes to make one complete swing from point A to C and back to A. It is measured in units of seconds.
What is the relationship between the period and frequency of a pendulum?
-The frequency of a pendulum is the reciprocal of its period. The period is the time taken for one complete cycle, while the frequency is the number of complete cycles that occur in one second, measured in hertz.
What formula is used to calculate the period of a simple pendulum, and what variables does it involve?
-The period of a simple pendulum is calculated using the formula T = 2π * √(L/g), where L is the length of the pendulum and g is the gravitational acceleration of the planet.
Why is the mass of the pendulum not considered in the period calculation?
-The mass of the pendulum is not part of the period calculation because the period of a simple pendulum is independent of the mass. This is due to the fact that the mass of the bob and the string are assumed to be negligible in the context of the pendulum's motion.
How does the length of the pendulum affect its period?
-As the length of the pendulum (L) increases, the period also increases because L is in the numerator of the fraction under the square root in the period formula. This means that a longer pendulum takes more time to complete a swing.
What is the relationship between gravitational acceleration and the period of a pendulum?
-The gravitational acceleration (g) is inversely related to the period of a pendulum. As gravitational acceleration increases, the period decreases because g is in the denominator of the period formula.
How can you determine the gravitational acceleration of an unknown planet using a simple pendulum?
-You can determine the gravitational acceleration of an unknown planet by knowing the length of the pendulum and the time it takes to complete a certain number of swings. Using the rearranged period formula, g = 4π² * L / T², you can solve for g given the values of L and T.
What is the significance of the period being two seconds in a grandfather clock pendulum?
-In a grandfather clock, the period being two seconds means that it takes two seconds for the pendulum to complete one full swing from the tick to the tock. This is because the one second interval mentioned is only half a cycle, from A to C, and not the full return trip to A.
Outlines
📐 Introduction to Simple Pendulum Dynamics
This paragraph introduces the concept of a simple pendulum, explaining the significance of a complete swing in determining the pendulum's period and frequency. The period, denoted by 'T', is the time taken for one complete swing from point A to C and back to A. Frequency, the reciprocal of the period, is the number of complete swings per second. The paragraph emphasizes the importance of understanding the relationship between period and frequency for solving pendulum-related problems. It also introduces the formula for the period of a pendulum, which depends on its length and the gravitational acceleration of the planet it's on. The formula is T = 2π√(L/g), where L is the length of the pendulum, and g is the gravitational acceleration. The paragraph concludes by noting that the period of a simple pendulum is independent of the mass of the pendulum bob.
🌕🌖 Calculating Pendulum Period and Frequency on Earth and the Moon
This paragraph demonstrates how to calculate the period and frequency of a simple pendulum with a given length on Earth and on the Moon. It starts by converting the length from centimeters to meters and then applies the period formula T = 2π√(L/g) using Earth's gravitational acceleration (9.8 m/s²). The calculated period on Earth is 1.679 seconds, and the frequency is found by taking the reciprocal of the period, resulting in 0.5956 Hz. For the Moon, where the gravitational acceleration is approximately 1.6 m/s², the period increases to 4.16 seconds due to the lower gravitational pull, and the frequency decreases to 0.24 Hz. The inverse relationship between gravitational acceleration and the period of a pendulum is highlighted, along with the direct relationship between frequency and gravitational acceleration.
⏱️ Determining Period and Frequency from Given Cycles
The paragraph explains how to determine the period and frequency of a pendulum when the number of cycles and the total time are known. Using the formula for period (T = time / number of cycles), the period is calculated to be 1.5 seconds for a pendulum that completes 42 cycles in 63 seconds. The frequency, the reciprocal of the period, is then calculated to be approximately 0.67 Hz. The process involves basic arithmetic operations and understanding of the relationship between time, cycles, period, and frequency. The paragraph also includes a step-by-step guide to finding the length of the pendulum using the rearranged period formula L = gT² / (4π²), given the period and Earth's gravitational acceleration.
🌐 Estimating Gravitational Acceleration of an Unknown Planet
This paragraph discusses a method to estimate the gravitational acceleration of an unknown planet using a simple pendulum. By knowing the length of the pendulum and the number of swings it makes in a given time, one can calculate the gravitational acceleration using the rearranged period formula g = 4π²L / T². The example provided calculates the gravitational acceleration of a planet where a pendulum with a length of 80 centimeters completes 28 swings in 45 seconds. The calculated gravitational acceleration is 12.2 m/s², which is then compared to Earth's gravitational acceleration to determine it is 1.24 times greater than that of Earth's, indicating a higher gravitational pull on this unknown planet.
⏳ Period of a Grandfather Clock's Pendulum
The paragraph focuses on the period of a pendulum used in a grandfather clock, which has a one-second interval between its tick and tock. It clarifies that the period of the pendulum is two seconds, as the one-second interval is only half of a complete swing. Using the formula for the length of a pendulum L = gT² / (4π²), and knowing that the period (T) is two seconds on Earth with a gravitational acceleration (g) of 9.8 m/s², the length of the pendulum is calculated to be 0.993 meters. This example illustrates how to apply the pendulum's period to find its length, which is a key aspect of understanding pendulum dynamics.
🌟 Period Variation with Gravitational Acceleration Change
This paragraph explores how the period of a pendulum changes when the gravitational acceleration changes, such as when moved from Earth to another planet with a different gravitational acceleration. It uses the ratio of the periods on two different planets to show that the new period (T2) can be calculated using the formula T2 = T1√(g1/g2), where T1 is the period on Earth, and g1 and g2 are the gravitational accelerations on Earth and the other planet, respectively. An example is provided where a pendulum with a period of 1.7 seconds on Earth is moved to a planet with a gravitational acceleration of 15 m/s², resulting in a new period of 1.37 seconds. This demonstrates the inverse relationship between gravitational acceleration and the period of a pendulum.
🔗 Mass Independence of Simple Pendulum's Period
The final paragraph addresses the misconception that the mass of a pendulum affects its period. It reaffirms that the period of a simple pendulum is independent of the mass of the pendulum bob, as the formula for the period does not include mass. Therefore, if the mass of the pendulum is doubled from m to 2m, the period remains unchanged. This is a crucial point to understand when analyzing the behavior of simple pendulums, as it simplifies the calculations and focuses on the length and gravitational acceleration as the primary determinants of a pendulum's period.
Mindmap
Keywords
💡Simple Pendulum
💡Period
💡Frequency
💡Gravitational Acceleration
💡Length of the Pendulum
💡Complete Swing
💡Formulas
💡Practice Problems
💡Independence from Mass
💡Reciprocal Relationship
Highlights
Introduction to the concept of a simple pendulum and its components.
Explanation of a complete swing of a pendulum and its relevance to determining period and frequency.
Definition of the period (T) of a pendulum as the time for one complete swing.
Definition of frequency as the reciprocal of the period and its unit in hertz.
Formula for calculating the period of a pendulum based on its length and gravitational acceleration.
Clarification that the period of a simple pendulum is independent of the mass of the bob.
Explanation of the relationship between the length of the pendulum and the period of its swing.
Discussion on how gravitational acceleration affects the period of a pendulum's swing.
Formula for calculating the frequency of a pendulum in terms of its length and gravitational acceleration.
Practical example of calculating the period and frequency of a pendulum on Earth and the Moon.
Step-by-step calculation of the period and frequency for a pendulum on Earth with a given length.
Calculation of the period and frequency for the same pendulum on the Moon with different gravitational acceleration.
Method to determine the gravitational acceleration of an unknown planet using a simple pendulum.
Calculation of the length of a pendulum given its period and gravitational acceleration on Earth.
Demonstration of how to find the gravitational acceleration of an unknown planet using a pendulum's period and known length.
Explanation of how the period of a grandfather clock's pendulum relates to its tick-tock cycle.
Calculation of the length of a pendulum used in a grandfather clock with a two-second period on Earth.
Method to determine the period of a pendulum on a planet with a different gravitational acceleration.
Illustration of how the period of a pendulum changes with different gravitational accelerations.
Conclusion that the period of a simple pendulum does not change with mass, demonstrated through a comparison of pendulums with different masses.
Transcripts
in this video we're going to talk about
the simple pendulum
so let's begin by drawing
our vertical line
and let's
draw a pendulum
let's put it at an angle
let's call this point a
b
and point c
now as the pendulum
moves from point a to point b
and then the point c
and then as it returns from c to a
that is one complete swing
now the reason why i need to know
what a complete swing is is because it
can help you to determine the period and
the frequency
of a simple pendulum
the period represented by capital t
is the time that it takes
to make one complete swing that's going
from a to c and then c to a
you can also calculate the period by
taking
the time and dividing by the number of
cycles or the number of complete swings
so the period is measured in units
of seconds
so it's the time
that it takes to make one complete swing
or one complete cycle
the frequency is the reciprocal of the
period
to calculate the frequency you could
take the number of cycles or complete
swings
and divide it by the time
the unit of frequency
is
the reciprocal of the second it's one
over seconds
or equivalently hertz
so make sure you understand that the
period is the time it takes
to make one complete cycle
whereas the frequency is the number of
cycles that occurs in one second
now make sure that you're writing this
down because you're going to use some of
these formulas later
when we work on some practice problems
now in addition to the formulas that we
have on the board
there's some other formulas that you may
want to add to your list
l is the length of
the pendulum
the period depends on the length of the
pendulum the period is equal to 2 pi
times the square root
of the length of the pendulum divided by
the gravitational acceleration
so g is the gravitational acceleration
of the planet
so the gravitational acceleration for
the earth
as you know
it's 9.8 meters per second squared
so the time it takes to make one
complete swing
depends on
the length of the pendulum and the
gravitational acceleration
as you can see
the mass is not part of that equation
therefore the period of a simple
pendulum is independent of the mass
if you increase the mass of the bob it's
not going to change the period of the
pendulum
so when dealing with a simple pendulum
we're assuming that the mass of the
string relative to the bob that it's
attached to can be ignored
but for a typical test that you might
take on this just know that the period
of a simple pendulum doesn't depend on
the mass of the bob
it only depends on these two things
the length and the gravitational
acceleration
now since l
is on top of that fraction
increase in l will increase the period
that means that
as you increase the length of the string
the time it takes for the bob to go from
a to c and then c to a and that time is
going to increase
it's going to take longer to make the
journey forward and then back
now if you increase the gravitational
acceleration let's say if you brought
the pendulum to a planet where
the gravity is stronger
the period is going to decrease
because g is in the denominator of that
fraction
there's an inverse relationship between
g and t
now to calculate the frequency you could
use this formula the frequency is one
over two pi
times the square root of g over l
so since l is on the bottom as you
increase l
the frequency
decreases
so the frequency the number of cycles
that occur in one second or the number
of complete swings that the pendulum
makes in a single second that's
inversely related to the length
of
the pendulum
but if you increase the gravitational
acceleration the frequency is going to
go up
so if the period goes up the frequency
goes down
and if the period goes down
the frequency goes up
so frequency and period they're
inversely related
now let's work on some practice problems
what is the period and frequency
of a simple pendulum that is 70
centimeters long on the earth and on the
moon
so let's begin with a picture
so here is our pendulum
and we were given the length of the
pendulum it's 70 centimeters long but we
want to convert that to meters
so we know that
100 centimeters is equal
to a meter
so to convert from centimeters to meters
simply divide by a hundred
and so the left is going to be point
seventy meters
what formula do we need in order to
calculate
the period and the
frequency in this problem to calculate
the period we could use this equation
it's 2 pi
times the square root
of the left over the gravitational
acceleration now on the earth we know
what the gravitational acceleration is
g is 9.8
so now we just got to plug in everything
into this formula so it's going to be 2
pi
times the square root
of 0.7 meters
divided by
9.8 meters per second squared
so looking at the units
the unit meters will cancel
and then when you take the square root
of
s squared is going to become s
eventually
and so the period is going to be in
seconds
now let's go ahead and plug this in
so the period for part a or
when the pendulum is on the earth
is going to be 1.679
seconds
now we need to calculate the frequency
the frequency is simply 1 over the
period
so it's 1 divided by
1.679 seconds
and you're gonna get point
five nine
five six
and then this is in it's gonna be
seconds to the minus one or hertz
so that's the frequency for the first
part that is when the pendulum is on the
earth
now let's move on to the next part
so what about if the pendulum
is on the moon
what will be the period
of the simple pendulum
now the gravitational acceleration on
the moon is about 1.6
meters per second squared
so everything is going to be the same
except the value of g
so g is going to be 1.6
instead of 9.8
so in the last problem
the period was i'm just going to write
it here
it was
1.679 seconds
that was on the earth
now the period on the moon let's call it
tm do you think it's going to be greater
than 1.679 seconds or less than
well as we go from the earth to the moon
the gravitational acceleration is
decreasing
and since that's on the bottom of the
fraction
it's inversely related to the period
that means when one goes up the other
goes down or when one goes down the
other goes up
so g is decreasing that means that the
period is going to increase
so on the moon it's going to take a
longer time to make a complete swing
so we should get an answer that's bigger
than 1.679 seconds
so let's go ahead and plug this into our
calculator
so this is going to be four
point one
six seconds
so as you can see
it's much bigger than 1.679 seconds
so as the gravitational acceleration
decreases the period is going to
increase they're inversely related
now let's calculate the frequency
so the frequency is going to be 1 over
the period
so that's 1 over 4.16 seconds
and that's going to be .24 hertz
so that's the frequency of the pendulum
on the moon
let's move on to our next problem
a pendulum makes 42 cycles in 63 seconds
what is the period and frequency of the
pendulum
the period is going to be
the time
divided by
the number of cycles
so we have 42 cycles occurring in a time
period of 63 seconds
so if we take 63 divided by 42
or 63 seconds divided by 42 cycles
we get that there's 1.5 seconds per
cycle
so the time that it takes to make one
complete cycle is 1.5 seconds
so thus we could say that the period is
1.5 seconds
so that's the answer for the first part
of part a
so now we can calculate the frequency
the frequency is 1 over the period
so it's 1 over 1.5 seconds
you're going to get 0.6 repeating which
we could round that to 0.67
and you could say the units are seconds
to the minus 1 hertz
so what this means is that
there's 0.67 cycles that are occurring
every second
so the units here
it's really technically one cycle
divided by
1.5 seconds
and so you get 0.67 cycles
per second
so that's what the frequency in hertz is
telling you is the number of cycles that
are occurring
every one second
now what about part b what is the length
of the pendulum on earth
so let's clear away a few things and
let's just rewrite
the first two answers that we had
so how can we calculate the length
of the pendulum when it's on the earth
well let's start with this formula t is
equal to 2 pi
times the square root
of l over g
we have the period and we know the
gravitational acceleration of the earth
we just need to isolate l in this
equation
so let's do some algebra
let's square both sides of the equation
so on the left it's going to be t
squared
2 pi squared is going to be
2 squared is 4 and then pi times pi is
pi squared
and then when we square root i mean when
we take the square of a square root
both the square and the square root will
cancel and so we're just going to get
l over g
now we need to get l by itself
so we're going to multiply both sides by
g
and then divide by 4 pi
squared
by doing this on the right side we can
cancel g and we can also cancel
4 pi squared
so we're just going to get l on the
right side
so we could say that l
is equal to everything that we see here
so the left of the pendulum
is going to be the gravitational
acceleration g
times the square of the period
divided by
four pi squared so that's the formula
that we can use to get the length
of the pendulum
now let's go ahead and plug everything
in
and let's fight the units as well
so g is going to be 9.8
meters per second squared
and then we're going to multiply that by
the square of the period so that's 1.5
seconds
squared
and then let's divide that by 4 pi
squared which doesn't contain units
so we can see that
second squared will cancel with s
squared here
and so l
is going to be in meters
so let's plug in 9.8 times 1.5 squared
divided by now you want to put this in
parenthesis otherwise your calculator
will divide by 4 and then multiply by pi
squared
and you'll get a different answer
so let's divide by four pi squared in
parenthesis
and you should get
point
five five
eight five meters
so this is the length of the pendulum on
earth
that has these features
if you are transported to an unknown
planet
where the gravity of that planet is
different from the earth
how can you determine the gravitational
acceleration of that planet
well this problem will help you to see
how
all you need is a simple pendulum
if you know the length of the pendulum
and how many swings that the pendulum
make in the given time period
you have all that you need
to calculate or even estimate
the gravitational acceleration of that
planet
so let's work on this problem
the first thing we need to do is
calculate the period
the period is going to be
the time
divided by the number of cycles
so this particular pendulum
makes 28 complete swings or 28 cycles
in a time period
of 45 seconds
so let's divide 45 seconds by 28 cycles
and this will give us the period which
is 1.607
seconds per cycle
so this tells us the time it takes to
complete
one cycle
so it takes
1.607 seconds to make one complete swing
so that's the period
now that we know the period
we could use this formula
to calculate the gravitational
acceleration
but we need to rearrange that formula
so like we did last time let's go ahead
and square
both sides of this equation
so we're going to get t squared is equal
to 4 pi squared
and the square root symbol will
disappear so it's going to be times l
over g
now we need to get g by itself
so let's multiply both sides by
g
over t squared
so the left side i'm going to multiply
by g over t squared
on the right side g is going to cancel
on the left side t squared is going to
cancel
so the only thing that we have on the
left side is g
so we have g is equal to
4 pi squared
times l and on the bottom we have t
squared
so this is the form of that we could use
to calculate the gravitational
acceleration
of any planet using
a simple pendulum all we need to know
is the length of the pendulum and the
period
or the time it takes to make one
complete swing
so for this problem it's going to be 4
pi squared
times the length of the pendulum which
is 80 centimeters or if you divide that
by 100 that's going to be 0.80 meters
and divided by the square of the period
the period is 1.607
seconds
and don't forget to square it
so the answer is
12.2
meters per second squared
so this is the gravitational
acceleration
of the planet
now how many g's
is this with respect to the earth
if we take that number
and divide it by the gravitational
acceleration of the earth
this is going to be 1.24
so it's 1.24 g's
so it's 1.24
times greater than the gravitational
acceleration of the earth
so that's what that figure means
whenever you see it
it simply compares the acceleration that
you're experiencing with the
acceleration of the earth
number four what is the length of a
simple pendulum used in a grandfather
clock
that has one second between its tick and
its talk on earth
so let's draw a picture
feel free to pause the video and try
this problem if you want to
so here is our simple pendulum
and let's label three points point a
b
and point c
actually let's get rid of that
so
it's going to take one second
for the grandfather clock to go from a
to c
where it's going to make
the first noise it's tick noise then
it's going to take another second for it
to go from c to a
where it's going to make the talk noise
so it's like tick tock tick tock and
just
oscillates between a and c
so we need to realize is that
the period is not one second
but two seconds
it takes two seconds to make a complete
swing
the one second is just
it's half of a cycle it's going from a
to c but doesn't include the return trip
so the period for grandfather clock is
two seconds
so with that information we can now
calculate the length
of the simple pendulum that is in the
grandfather clock
and so we're going to use this formula l
is equal to
g t squared
divided by
4 pi squared
so on earth g is 9.8
the period for the grandfather clock is
2 seconds
and then divided by 4 pi squared and
let's put that in parenthesis as well
and
i almost forgot to square the period so
don't make that mistake
so the answer is going to be .993
meters
so that that's the length of the
grandfather clock
given a period
of two seconds
number five
a certain pendulum has a period of 1.7
seconds on earth
what is the period of this pendulum on a
planet that has a gravitational
acceleration
of 15 meters per second squared
well in order to calculate the period we
can use this formula
it's 2 pi
times the square root of l over g
but let's write down what we know in
this problem and what we need to find
so the period on the earth let's call it
t1
that's 1.7 seconds
now we know that the gravitational
acceleration on the earth
we'll call that g1 is 9.8 meters per
second squared
we wish to calculate the new period
on some other planet
t2
and we're given the gravitational
acceleration
of that planet which is 15 meters per
second squared
so we have t1 and g1
how can we calculate t2 if we know g2
well let's make a ratio of this equation
we're going to divide t2 by t1
so if t is equal to what we see here
t2 is going to be
2 pi
square root
l
over g2
now the reason why i didn't write l2
over g2 is because l doesn't change
we're dealing with the same pendulum
that was on the earth that is now in
this new planet
so therefore l is constant we don't need
to change the subscript for l
only g and t changes
so t one
is going to be
2 pi times the square root of l
over g1
so we can cross out 2 pi
and we can cancel l
and so we're left with t2 over t1 is
equal to the square root of 1 over g2
divided by the square root of 1
over g1
now let's multiply the top and the
bottom by the square root of g1
doing so
will give us some one in the denominator
so it's going to be
the square root of this is one
times g one
so we'll have g1 on the top
g2 is gonna stay on the bottom
and then this simply becomes one
the square root of one is one
so we could say that t2 over t1
is equal to the square root of g1 over
g2
and then finally we can multiply both
sides by t1
to cancel this
so we have this equation
the second period t2 is going to equal
the first period
times the square root of g1 over g2
now let's plug in some values t1 is 1.7
g1 is 9.8
g2 is 15.
so 1.7 times the square root of 9.8 over
15
that's going to be
1.37
seconds
so that's the new period
so as we could see we increase the
gravitational acceleration
from
9.8 to 15.
and because g is on the bottom it's
inversely related to t
so as we increase g
the period decreased it went from 1.7
to 1.37 seconds
so these two are inversely related
number six
the period of a simple pendulum with
mass m is t
which of the following expressions
represent the period of a simple
pendulum with mass 2m
well if we write the formula for the
simple pendulum
notice that it doesn't depend on the
mass
so if we change the mass from m
to 2m it's not going to change it period
the period was initially t
it's going to remain t
so for a simple pendulum the period is
independent
from the mass it doesn't change with the
mass
so the answer is gonna be d there's no
change
you
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