The Simple Pendulum - Free Educational videos for Students in K-12

The Simple Pendulum - By The Organic Chemistry Tutor

Transcript


00:0-1	in this video , we're going to talk about the
00:02	simple pendulum . So let's begin by drawn our vertical
00:06	life and let's dry pendulum . Let's put out an
00:12	angle . Let's call this point A . B .
00:17	And point C . Now as the pendulum moves from
00:24	point A . To point B . And then the
00:26	point C . And then as it returns from C
00:30	. To A . That is one complete swing .
00:36	Now the reason why I need to know what a
00:38	complete swing is is because it can help you to
00:41	determine the period and the frequency of a simple pendulum
00:47	. The period represented by capital T . Is the
00:51	time that it takes to make one complete swing that's
00:55	going from A . To C . And then see
00:56	to A . You can also calculate the period by
01:00	, taken the time and divided by the number of
01:03	cycles or the number of complete swings . So the
01:08	period is measured in units of seconds . So it's
01:15	the time that it takes to make one complete swing
01:18	or one complete cycle . The frequency is the reciprocal
01:24	of the period . To calculate the frequency , you
01:27	could take the number of cycles or complete swings and
01:30	divided by the time the unit of frequency is the
01:37	reciprocal of the second . It's one of the seconds
01:40	or equivalently hurts . Mhm . So make sure you
01:46	understand that the period is the time it takes to
01:49	make one complete cycle . Whereas the frequency is the
01:54	number of cycles that occurs in 1/2 . Now ,
01:57	make sure that you're writing this down because you're going
01:59	to use some of these formulas later when we work
02:02	on something practice problems . Mhm . Now , in
02:05	addition to the former is that we have on the
02:07	board , there's some other formulas that you may want
02:10	to add to your list . L . Is the
02:14	left of the pendulum . Mhm . The period depends
02:19	on the left of the pendulum . The period is
02:22	equal to two pi times the square root of the
02:26	left of the pendulum , divided by the gravitational acceleration
02:30	. So G . Is the gravitational acceleration of the
02:32	planet . So the gravitational acceleration for the Earth ,
02:36	as you know , it's 9.8 m/s squared . So
02:42	the time it takes to make one complete swing depends
02:47	on the left of the pendulum and the gravitational acceleration
02:51	. As you can see , the mass is not
02:54	part of that equation . Therefore the period of a
02:57	simple pendulum is independent of the mass . If you
03:01	increase the mass of the bob , it's not going
03:03	to change the period of the pendulum . So when
03:07	dealing with a simple pendulum , we're assuming that the
03:10	mass of the string relative to the bob that it's
03:13	attached to can be ignored . But for a typical
03:17	test that you might take on this , just know
03:19	that the period of a simple pendulum doesn't depend on
03:23	the mass of the body . It only depends on
03:25	these two things , the less than the gravitational acceleration
03:29	. Now , since L . Is on top of
03:32	that fraction increase in L . Will increase the period
03:39	. That means that as you increase the length of
03:42	the string , the time it takes for the bob
03:46	to go from A . To C . And then
03:47	see the A . And that time is going to
03:49	increase . It's going to take longer to make the
03:53	journey forward and then back . Now if you increase
03:57	the gravitational acceleration , let's say if you brought the
04:00	pendulum to a planet where the gravity is stronger ,
04:05	the period is going to decrease because she is in
04:09	the denominator of that fraction . There's an inverse relationship
04:12	between G . And T . Now To calculate the
04:20	frequency , you could use this formula . The frequency
04:23	is 1/2 pi times the square root of G over
04:27	out . So since Ellis on the bottom as you
04:30	increase l the frequency decreases . So the frequency the
04:37	number of cycles that occur in one second where the
04:41	number of complete swings that pendulum makes in a single
04:43	second . That's inversely related to the left of this
04:49	, the pendulum . But if you increase the gravitational
04:53	acceleration , the frequency is going to go up .
04:59	So if the period goes up the frequency goes down
05:03	and if the period goes down the frequency goes up
05:07	, so frequency and period they're inversely related . Now
05:11	let's work on some practice problems . What is the
05:15	period and frequency Of a simple pendulum ? That is
05:18	70 cm long . On the Earth and on the
05:21	moon ? Yeah . So let's begin for picture .
05:28	Mhm . So here is our pendulum . Yeah .
05:33	And were given the length of the pendulum , it's
05:36	70 cm long . But we want to convert that
05:38	to meters . So we know that 100 cm is
05:45	equal to a meeting . So to convert from cm
05:49	to m simply divide by a 100 and so the
05:52	left is going to be point 70 m . What
06:00	formula do we need in order to calculate the period
06:03	and frequency in this problem ? To calculate the period
06:07	. We could use this equation . It's two pi
06:10	times the square root . Yeah of the left over
06:17	the gravitational acceleration . Now on the Earth we know
06:20	what the gravitational acceleration is . G . s 9.8
06:27	. It's not we just gotta plug in everything into
06:29	this formula . So it's gonna be two pi times
06:31	the square root of .7 m , divided by 9.8
06:37	m/s squared . So looking at the units the unit
06:42	meters will cancel . And then when you take the
06:45	square root of S squared is going to become S
06:49	. Eventually . And so the period is going to
06:51	be in seconds . Now let's go ahead and plug
06:55	this in . So the period for part A or
07:06	when the pendulum is on the earth , It's going
07:09	to be 1.679 seconds . Now we need to calculate
07:15	the frequency . The frequency is simply one over the
07:18	period . So it's one divided by 1.67 , 9
07:23	seconds and you're going to get points 59 56 And
07:39	then this is in , it's going to be second
07:41	to the -1 or hurts . So that's the frequency
07:44	for the first part . That is when the pendulum
07:47	is on the Earth . Now let's move on to
07:50	the next part . Yeah . So what about if
07:54	the pendulum is on the moon ? What will be
07:59	the period of the simple pendulum ? Now the gravitational
08:03	acceleration on the moon is about 1.6 meters per second
08:08	squared . So everything is gonna be the same except
08:12	the value of G . Yes , So G is
08:20	going to be 1.6 Instead of 9.8 . It's in
08:26	the last problem . The period was I'm just going
08:31	to write it here , It was 1.679 seconds .
08:37	Yeah , that was on the Earth . Now ,
08:40	the period on the moon , let's call it .
08:42	TM do you think it's going to be greater than
08:44	1.679 seconds or less than well as we go from
08:50	the Earth to the moon ? The gravitational acceleration is
08:53	decreasing . And since that's on the bottom of the
08:57	fraction , it's inversely related to the period . That
09:01	means when one goes up the other goes down or
09:03	when one goes down the other goes up so G
09:06	. Is decreasing . That means that the period is
09:08	going to increase , so on the moon it's going
09:11	to take a longer time to make a complete swing
09:15	. So we should get an answer that's bigger than
09:17	1.679 seconds . So let's go ahead and plug this
09:21	into our calculator . So this is going to be
09:30	four 0.1 six seconds . Yeah . Yeah . So
09:41	as you can see It's much bigger than 1.679 seconds
09:46	. So as the gravitation of celebration decreases , the
09:48	period is going to increase their inversely related . Now
09:52	let's calculate the frequency . So the frequency is gonna
09:55	be one over the period . So that's one over
09:58	4.16 seconds . And that's gonna be .24 Hz .
10:09	So that's the frequency of the pendulum on the moon
10:13	. Let's move on to our next problem . A
10:16	Pendulum makes 42 cycles in 63 seconds . What is
10:20	the period and frequency of the pendulum ? The period
10:28	is going to be the time divided by the number
10:32	of cycles . So we have 42 cycles occurring in
10:40	the time period of 63 seconds . So if we
10:45	take 63 divided by 42 or 63 seconds divided by
10:50	42 cycles , We get that there's 1.5 seconds per
10:55	cycle . So the time that it takes to make
10:59	one complete cycle is 1.5 seconds . So that we
11:02	could say that the period Is 1.5 seconds . Yeah
11:09	, so that's the answer for the first part of
11:11	party . So now we can calculate the frequency ,
11:14	The frequency is one over the period . So it's
11:17	one over 1.5 seconds . You're gonna get .6 repeat
11:26	in which we profound that 2.67 . And you can
11:30	say the units are second to the -1 hurts .
11:34	So what this means is that There's .67 cycles that
11:40	are occurring every second . So the units here ,
11:45	it's really technically one cycle divided by 1.5 seconds .
11:51	And so you get points 67 cycles per second .
11:57	So that's what the frequency in hertz is telling you
12:01	, is the number of cycles that are occurring every
12:03	one second Now . What about part B ? What
12:08	is the length of the pendulum on Earth ? Okay
12:14	, so let's clear away a few things and let's
12:17	just rewrite . Mhm . The first two answers that
12:20	we had . So how can we calculate the length
12:27	of the pendulum when it's on the earth ? Well
12:31	, let's start with this formats . He is equal
12:33	to two pi times the square root of L .
12:37	A V . G . We have the period and
12:40	we know the gravitational acceleration of the Earth . We
12:43	just need to isolate L . In this equation .
12:45	So let's do some algebra . Let's square both sides
12:50	of the equation . So on the left it's going
12:52	to be t squared . two Pi squared is going
12:55	to be two squared is four . And then pied
12:58	some supplies , pi squared . And then when we
13:01	square root , I mean when we take the square
13:03	of a square root , both the square and the
13:06	square , we will cancel . And so we're just
13:08	going to get L O V E G . Mhm
13:13	. Now we need to get all by itself .
13:16	So we're going to multiply both sides by G .
13:18	And then divide by four pi squared by doing this
13:25	on the right side . We can cancel G .
13:27	And we can also cancel four pi squared . So
13:31	we're just going to get out on the right side
13:34	. So we can say that L . Is equal
13:36	to everything that we see here . So the less
13:39	of the pendulum , it's going to be the gravitational
13:42	acceleration G times the square of the period , divided
13:47	by yeah , four pi squared . So that's the
13:51	form of that . We can use to get the
13:52	length of the pendulum . Now let's go ahead and
13:58	plug everything in . Let's fight a unit as well
14:02	. So G is going to be 9.8 meters per
14:05	second squared . And then we're gonna multiply that by
14:09	the square of the period . So that's 1.5 seconds
14:13	squared . And then let's divide that by four pi
14:16	squared , which doesn't contain units . So we can
14:20	see that second squared will cancel with S squared here
14:24	. And so L . Is going to be in
14:27	meters . So let's plug in 9.8 times 1.5 squared
14:34	divided by Now . You want to put this apprentices
14:36	. Otherwise your calculated will divide by four and then
14:38	multiply by pi square and you'll get a different answer
14:43	. So let's divide by four pi squared in parentheses
14:46	and you should get point 55 85 m . So
14:56	this is the left of the pendulum on Earth that
14:59	has these features . If you are transported to an
15:04	unknown planet where the gravity of that planet is different
15:08	from the earth , how can you determine the gravitational
15:13	acceleration of that planet ? Well , this problem will
15:16	help you to see how all you need is a
15:19	simple pendulum . If you know the left of the
15:21	pendulum and how many swings that the pendulum make in
15:25	the given time period . You have all that you
15:28	need to calculate or even estimate the gravitational acceleration of
15:33	that planet . So let's work on this problem .
15:38	The first thing we need to do is calculate the
15:39	period , The period is going to be the time
15:45	divided by the number of cycles . So this particular
15:53	pendulum Makes 28 complete swings or 28 cycles in a
16:00	time period Of 45 seconds . So let's divide 45
16:05	seconds by 28 cycles . And this will give us
16:09	the period Which is 1.607 seconds per cycle . So
16:16	this tells us that time it takes to complete one
16:19	cycle . So it takes 1.607 seconds to make one
16:24	complete swing . So that's the period now that we
16:28	know the period . We could use this formula to
16:34	calculate the gravitational acceleration , but we need to rearrange
16:38	that formula . So like we did last time ,
16:41	let's go ahead and square both sides of this equation
16:45	. So we're going to get T squared is equal
16:48	to four pi squared , and the square root symbol
16:51	will disappear . So it's gonna be times L over
16:54	G . Now we need to get G by itself
17:00	. So let's multiply both sides by G over T
17:04	squared . So the left side , I'm going to
17:07	multiply by G over T squared . Mhm . On
17:11	the right side , G is going to cancel .
17:14	On the left side , T squared is going to
17:15	cancel . So the only thing that we have on
17:18	the left side is G . So we have G
17:20	is equal to four pi squared times L . And
17:26	on the bottom we have T squared . So this
17:28	is the form of that we could use to calculate
17:31	the gravitational acceleration of any planet using a simple pendulum
17:36	. All we need to know is the left of
17:38	the pendulum , end of period or the time it
17:41	takes to make one complete swing . So for this
17:45	problem is going to be four pi squared times the
17:48	left of the pendulum , which is 80 centimeters .
17:50	Or if you divide that by 100 that's gonna be
17:53	1000.80 m And divided by the square of the period
17:58	. The period is 1.607 seconds . Mhm . And
18:06	don't forget to square . Yeah . So the answer
18:17	is 12.2 meters per second squared . So this is
18:24	the gravitational acceleration of the planet . Now , how
18:31	many gs is this with respect to the Earth ?
18:36	If we take that number and divided by the gravitational
18:40	acceleration of the Earth ? Yeah , This is going
18:46	to be 1.24 . So it's 1.24 Gs . So
18:51	it's 1.24 times greater than the gravitational acceleration of the
18:55	earth . So that's what that figure means whenever you
19:00	see it , it simply compares the acceleration that you're
19:04	experiencing with the acceleration of the Earth number four .
19:09	What is the length of a simple pendulum used in
19:11	a grandfather clock that has one second between its tick
19:15	and it's talk on earth . So let's try a
19:18	picture . Mhm . Feel free to pause the video
19:21	and try this problem if you want to . So
19:25	here is our simple pendulum . And let's label three
19:28	points point a . B . And point C .
19:36	Mhm . Actually get rid of that . So It's
19:40	going to take 1 2nd for the grandfather clock to
19:43	go from A . To C . Where it's going
19:45	to make the first noise , it's tick noise .
19:49	Then it's going to take another second for it to
19:51	go from C . To A . Where it's going
19:53	to make the talk noise . So it's like tic
19:55	tac tic tac and just oscillates between A . And
19:59	C . So we need to realize is that The
20:02	period is not 1 2nd , but two seconds .
20:06	It takes two seconds to make a complete swing .
20:09	The 1 2nd is just , it's half of a
20:13	cycle . It's going from A to C . But
20:15	doesn't include the return trip . So the period for
20:19	grandfather clock is two seconds . So with that information
20:23	, we can now calculate the length of the simple
20:26	pendulum that is in their grandfather clock . And so
20:31	we're going to use this formula L . Is equal
20:33	to GT squared divided by four pi squared . Yeah
20:41	. Mhm . So on Earth G . is 9.8
20:45	. The period for the grandfather clock is two seconds
20:49	and then divided by four pi squared . And let's
20:52	put that in parentheses as well . And I almost
21:05	forgot to square the period . So don't make that
21:08	mistake . So the answer is going to be point
21:16	993 m . So that that's the length of the
21:20	grandfather clock . Given a period of two seconds ,
21:24	number five , A certain pendulum has a period of
21:28	1.7 seconds on Earth . What is the period of
21:32	this pendulum on a planet that has a gravitational acceleration
21:36	of 15 m/s squared ? Well , in order to
21:41	calculate the period , we can use this formula ,
21:44	it's two pi times the square root of L over
21:47	G . But let's write down what we know in
21:50	this problem and what we need to find . So
21:53	the period on the earth , let's call it t
21:55	. one . That's 1.7 seconds . Now we know
22:00	that the gravitational acceleration on the Earth , We'll call
22:03	that G . one is 9.8 meters per second squared
22:08	. We wish to calculate the new period on some
22:11	other planet T . Two and were given the gravitational
22:15	acceleration Of that planet , which is 15 m/s squared
22:20	. So we have T . One N . G
22:22	one . How can we calculate T . Two if
22:24	we know G . Two . Well let's make a
22:28	ratio of this equation . We're going to divide teach
22:31	you by T . one . So if T .
22:34	Is equal to what we see here T . two
22:37	is going to be two pi square root L over
22:42	G . Two . Now the reason why I didn't
22:45	write L . Two of G two is because L
22:47	. Doesn't change . We're dealing with the same pendulum
22:51	that was on the earth that is now in this
22:53	new planet . So therefore Ellis constant . We don't
22:56	need to change the sub script for al only GMT
23:00	changes . So it's the one it's going to be
23:06	two pi times the square root of L . Over
23:09	G . Mark . Mhm . So we can cross
23:12	out to pie and we can cancel out . Yeah
23:17	. and so we're left with T . two over
23:19	T one is equal to The square root of one
23:23	over G two Divided by the square root of one
23:26	over G . One . Now let's multiply the top
23:30	and the bottom by the square root of G .
23:32	one . Doing so well give us someone in the
23:40	denominator . So it's going to be The square root
23:44	of this is one times G one . So we'll
23:48	have G1 on the top . G two is going
23:50	to stay in the bottom And then this simply becomes
23:54	one . The square root of one is 1 .
23:57	So we could say that T two over T one
24:00	is equal to the square root of G . One
24:02	over G . Two . And then finally we can
24:07	multiply both sides by T . one to cancel this
24:13	. So we have this equation . The second period
24:17	T two is going to equal the first period Times
24:20	The Square Root of G . one Over G two
24:24	. Yeah . Now let's plug in some values .
24:30	T one is 1.7 . G one is 9.8 .
24:35	G two is 15 So 1.7 times the square root
24:43	of 9.8 over 15 , That's going to be 1.37
24:50	seconds . So that's the new period . So as
24:56	we could see , we increase the gravitational acceleration from
25:01	9.8-15 and because Gs on the bottom , it's inversely
25:05	related to T . So as we increase G The
25:09	period decreased , it went from 1.7 To 1.37 seconds
25:15	. So these two are inversely related Number six .
25:20	The period of a simple pendulum with mass m .
25:23	s.t . Which of the following expressions represent the period
25:27	of a simple pendulum with mass to em . Well
25:33	, if we write the formula for the simple pendulum
25:37	, notice that it doesn't depend on a mass ,
25:40	so if we change the mass from EM 22 M
25:44	, it's not going to change your period . The
25:46	period was initially t it's going to remain team .
25:50	So for a simple pendulum , the period is independent
25:55	from the mass . It doesn't change with the mass
25:59	, so the answer is going to be deep .
26:00	There's no change .

Summarizer

DESCRIPTION:

This physics video tutorial discusses the simple harmonic motion of a pendulum. It provides the equations that you need to calculate the period, frequency, and length of a pendulum on Earth, the Moon, or another planet. This video contains plenty of examples and practice problems.

OVERVIEW:

The Simple Pendulum is a free educational video by The Organic Chemistry Tutor.

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