The Simple Pendulum

Summary

This 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

00:00

📐 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.

05:02

🌕🌖 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.

10:06

⏱️ 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.

15:06

🌐 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.

20:07

⏳ 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.

25:08

🌟 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.

Keywords

💡Simple Pendulum

A simple pendulum is a weight, often referred to as the 'bob,' suspended from a fixed point so that it can swing freely back and forth under the influence of gravity. In the video, the simple pendulum is used to illustrate fundamental concepts of physics such as period and frequency. The pendulum's motion is described as swinging from point A to B to C and back to A, which constitutes one complete swing.

💡Period

The period of a pendulum, denoted by the symbol T, is the time it takes to complete one full cycle of its swinging motion, from one extreme position to the other and back to the starting point. The video explains that the period is crucial for understanding pendulum motion and is measured in seconds. It's used in the context of calculating the frequency and is shown to be independent of the pendulum's mass.

💡Frequency

Frequency is the reciprocal of the period and is a measure of how many complete oscillations occur per unit time. It is typically measured in Hertz (Hz), which is one cycle per second. The video script uses the frequency to demonstrate the inverse relationship with the period, where an increase in the period results in a decrease in frequency and vice versa.

💡Gravitational Acceleration

Gravitational acceleration, symbolized as 'g', is the acceleration that an object experiences due to the gravitational force of the Earth or another planet. In the video, it's mentioned that the gravitational acceleration on Earth is 9.8 meters per second squared. This value is used in the formulas to calculate the period of a pendulum and is shown to affect the period directly.

💡Length of the Pendulum

The length of the pendulum, denoted as 'l', is the distance from the point of suspension to the center of mass of the pendulum's bob. The video explains that the period of a simple pendulum is dependent on this length, with longer pendulums having longer periods. This relationship is used to calculate the period using the formula T = 2π√(l/g).

💡Complete Swing

A complete swing of a pendulum refers to one full back-and-forth motion from its rest position, passing through the lowest point and returning to the starting point. The video emphasizes that understanding what constitutes a complete swing is essential for determining the period and frequency of the pendulum's motion.

💡Formulas

The video script introduces several key formulas related to pendulum motion, including the formula for the period T = 2π√(l/g) and the formula for frequency f = 1/T. These formulas are central to the video's educational content, providing viewers with the mathematical tools to calculate and understand pendulum dynamics.

💡Practice Problems

Practice problems are used throughout the video to apply the concepts and formulas discussed. These problems involve calculating the period and frequency of pendulums on Earth and the Moon, determining the gravitational acceleration of an unknown planet, and finding the length of a pendulum given its period. They serve to reinforce learning and demonstrate practical applications of the concepts.

💡Independence from Mass

The video clarifies that the period of a simple pendulum is independent of the mass of the bob. This is an important concept as it contrasts with other physical systems where mass might significantly affect motion. The script uses this to explain that increasing the mass of the bob does not change the period of the pendulum.

💡Reciprocal Relationship

The video explains the reciprocal relationship between period and frequency, and between gravitational acceleration and period. This relationship is fundamental to understanding pendulum motion and is illustrated through the formulas and examples provided. For instance, if the gravitational acceleration increases, the period decreases, and if the length of the pendulum increases, the frequency decreases.

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