This course is designed for high school students preparing to take the AP* Physics 1 Exam. * AP Physics 1 is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product.

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Preparing for the AP Physics 1 Exam

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This course is designed for high school students preparing to take the AP* Physics 1 Exam. * AP Physics 1 is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product.

From the lesson

Waves and Sound

Topics include mechanical waves, sound, simple harmonic motion and applications of topics. You will watch 3 videos, complete 3 sets of practice problems and take 2 quizzes. Please click on the resources tab for each video to view the corresponding student handout.

- Dr. Paige K. EvansClinical Associate Professor

Math - Mariam ManuelInstructional Assistant Professor

University of Houston

When a spring is stretched from its equilibrium position,

Â it pushes with a force to restore the spring back to its original length.

Â The strength of this force is described by Hook's Law, as seen here.

Â F = -kx.

Â Notice that the force from a spring has a linear relationship to the stretch of

Â the spring.

Â We call this a linear restoring force.

Â The further a spring is stretched, the more force a spring applies.

Â A common confusion with this equation is the negative sign.

Â Remember that force and displacement are vectors.

Â And that positive and negative signs represent direction.

Â When a spring is pushed in a positive direction from it's equilibrium,

Â it will apply a force in the negative direction.

Â This is why a negative sign must appear in the equation.

Â A spring with an equilibrium length of .2 meters and

Â spring constant of 35 newton per meters is hung vertically.

Â A mass of 500 grams is attached to the end of the spring causing it to

Â stretch as it hangs from rest.

Â In this question I'm told that there is a mass of 500 gram that's added

Â vertically to cause a stretch in this spring.

Â So for instance, this spring may have had an original

Â length over here, which they give me is the 0.2.

Â And, now I've got this additional stretch that's added U to this mass over here.

Â So let's go ahead and solve for that.

Â Now, the forces that are acting over here,

Â I've got, the force on this spring, the restoring force,

Â that's acting in the opposite direction of this motion, is trying to pull it up.

Â And then I have the mg, the weight of this mass that is pulling the spring downwards.

Â So, the sum of my forces equaled 0.

Â This whole system's at rest right now, there's no acceleration.

Â I've got a restoring force minus mg equaling 0.

Â Restoring force equals kx.

Â And from here, I can go ahead and solve for x.

Â As mg over K.

Â X equals.

Â And be careful to convert.

Â It was 500 grams so I'm going to use 0.5 kilograms.

Â G's gravitational acceleration for which I can use 10.

Â K is 35, that's given to me in the question.

Â This gives me an x of 0.14 meters.

Â Now, this is the amount of stretch that occurs.

Â Had the question just asked me how much this spring stretched, that would be it.

Â I would stop there.

Â But the question wants to know what this total length of the spring is.

Â That means that I need to go ahead and add.

Â So for my total length, I'm going to add the stretch,

Â .14, to the original length, which was .2.

Â And this gives me .34 meters.

Â And that is my final answer.

Â >> A mass on a spring is a great example of a simple harmonic oscillator.

Â A simple harmonic oscillator is an object that exhibits periodic motion

Â that can be described accurately with sin and cosine functions.

Â Let's look at an example.

Â Suppose that a mass is placed on a frictionless horizontal surface, and

Â attached to a massless frame at its equilibrium position.

Â The mass is now pulled a distance positive x from equilibrium and released from rest.

Â The mass will experience a force in the negative direction due to the spring's

Â restoring force and will begin to accelerate in that direction.

Â As the spring gets closer to the equilibrium position, the mass speeds up.

Â But the force on the mass becomes less as the stretch decreases.

Â At the equilibrium position, the mass has a maximum speed.

Â But no force at all acts upon it.

Â The mass continues to move and is now displaced in the negative x direction.

Â The force from the string now acts in a positive direction.

Â The mass again begins to slow down until it is at rest for just an instant.

Â And again starts to move back in the positive direction.

Â At this end point where it turns around, the force on the mass and

Â therefore the acceleration experienced by the mass are at a maximum.

Â This periodic motion continues as the mass oscillates back and forth.

Â >> Since the system has a linear restoring force,

Â it fits the definition of what we call a SHO, or simple harmonic oscillator.

Â That means we can describe the objects motion using sin and cosine as seen here.

Â When using these equations make sure that your calculator is in radian mode.

Â A represents amplitude.

Â F represents the frequency of the object's motion.

Â This equation can be used to calculate the location of the mass at any point in time.

Â So when should you use sin and when should you use cosine?

Â Well, recall that sin of 0 is 0, which means that when x equals 0 at t equal 0,

Â the sin function is the best choice.

Â In this scenario the object starts in its equilibrium position.

Â If the object instead starts at its amplitude so

Â that x equals A at t equals A, cosine is the better choice.

Â Let's look at an example.

Â In this question, I need to set up an equation that allows me to solve for

Â the location of this mass at any point in time.

Â Now, here's something important to consider.

Â Am I going to use sin or cosine?

Â Well, according to the question, it tells me that t = 0.

Â This mass is stretched to that 0.5 meters from equilibrium.

Â So that means that it's not starting from equilibrium which means

Â I'm going to use cosine.

Â Then I'm going to start with my equation x equals

Â A cosine 2 pi f multiplied by t.

Â Now in this case, I don't have f.

Â I don't have the frequency.

Â But they tell me that the period of the motion of the oscillator is 1.2 seconds.

Â And remember there's that inverse relationship between frequency and period.

Â So I can re-write this as x equals A cosine,

Â 2 pi, 1 over T multiplied by any given

Â point in time, the lower case t.

Â Well, now let's consider this.

Â Is stretch to this maximum displacement over here of .5 meters, well,

Â in this case, because I'm at t = 0, this represents my amplitude.

Â So, I can actually go ahead and plug that value in.

Â So x would be that point that I'm solving for in this case,

Â where whatever position it may be at a certain time.

Â The amplitude is 0.5 cosine 2 pi one over 1.2 and

Â then t is any given point and time that I'm stalling for.

Â So I can leave this equation as here.

Â I can also simplify this part over here a little bit further.

Â If you'd like to, you can go ahead and do that.

Â But this part over here allows us to give an equation that will give you

Â the mass' location at any point in time.

Â >> The period of a mass oscillating on a massless spring can be calculated using

Â the equation seen here.

Â This relationship applies for a vertical or horizontal spring configuration.

Â In this equation, the mass attached to the spring in represented by the variable m

Â and the variable k represents the spring strength where spring constant.

Â A pendulum that swings an angle smaller than about 15 degrees is also a good

Â approximation of a simple harmonic oscillator.

Â The period of a pendulum can be calculated using the equation seen here.

Â In this similar looking equation the length of the pendulum is represented by

Â the variable L.

Â And g is the acceleration due to gravity.

Â For us, 10 meters per second squared.

Â Notice that the amplitude of motion does not change the period of either

Â oscillator.

Â As each is pulled further from equilibrium, there is more restoring force

Â and the oscillator will travel with a greater average speed.

Â So in this problem we have a speaker that is moving up and

Â down because it vibrates back and forth.

Â And that membrane that holds that speaker there is elastic in nature, so

Â we're going to treat this as an oscillator.

Â With an elastic nature, that means it has a linear restoring force.

Â And it wants us to calculate the max acceleration of that speaker as

Â it vibrates back and forth.

Â All I know that this amplitude here, and be careful with your choice of variables.

Â Little a is already used in this problem to talk to an acceleration.

Â So typically, we use a capital A to represent amplitude.

Â The amplitude in this problem is .004 meters.

Â Also, they told us the frequency of the speaker.

Â The speaker is making a note that is 20 hertz.

Â Very low, right at the borderline of human hearing.

Â And if I'm looking for a max acceleration, well,

Â it sounds like I might want to use Newton's second law.

Â The sum of all the forces on the speaker will equal its mass times its

Â acceleration.

Â So there's a mass here for this speaker, this magnet that's vibrating up and

Â down that we don't know,

Â but that's okay maybe we can get around that somehow by using another equation.

Â So if I were to solve this for

Â acceleration, I now have whatever force for this membrane, that's moving it up and

Â down, divided by the mass, that we don't know, should equal acceleration.

Â And we're out of luck for both at the moment.

Â So let's think about that also.

Â If this is elastic along these edges, and

Â we're considering this something like a spring F would equal kx.

Â Our linear restoring force times the spring constant.

Â Well I know that I have the largest force and

Â therefore the largest acceleration when I'm stretched the most.

Â When my object is all the way at the amplitude.

Â So that means I want the largest force when my x equals the amplitude a.

Â So I can sub that in up here.

Â So that I now have something that looks like this.

Â KA over m equals the max acceleration.

Â Now I don't know K and I don't know m.

Â But I do notice something in particular.

Â That looks familiar.

Â We have an equation that talks about the period of a mass on a spring.

Â And it has an m divided by k in that equation.

Â Maybe we can use that equation to sub in and solve in this one.

Â So I'm going to go ahead and scroll down to give myself some more room.

Â And let's take a look at the equation that we know.

Â The period of a mass on a spring is equal to 2 pi, the square root of m over k.

Â And I want to manipulate this so that I can solve this for k divided by m and

Â sub it into that equation in the top right.

Â So why don't I do some algebra first?

Â I can divide by 2 pi to move it to the left-hand side of the equation and

Â then square both sides, giving me an expression that looks like this.

Â Well, I don't want to solve for m over k, I want to solver for k divided my m.

Â So if I take the inverse of all this, or think about it as cross-multiplying,

Â I'm not looking at an equation that looks like this: k over m would equal

Â 2 pi over the period, and all of that is squared.

Â Again, I have an unknown.

Â They never told me the period of the oscillation, but

Â they did tell me the frequency.

Â And I know that 1 over the frequency equals the period.

Â They're inverses of one another.

Â So, again, I'm going to rearrange this equation one more time,

Â k over m equals 2 pi f squared.

Â That's what I'm going to take and sub in for those two terms, those two variables.

Â So, changing colors so that it becomes a little bit clearer.

Â K divided by m equals 2 pi f squared times

Â the amplitude will give us our maximum acceleration.

Â Subbing in my numbers for that.

Â The frequency was 20 hertz.

Â All of that squared.

Â The amplitude .004.

Â The acceleration, the max acceleration that this

Â beaker experiences, 63.17 m/s squared.

Â Okay, so

Â this tells me that I have a pendulum vibrating with a period of 2 seconds.

Â That means it repeats its motion every 2 seconds.

Â And it also tells me the amplitude.

Â And it wants to know where is this pendulum located

Â at very specific points in time.

Â Before I get into any equations,

Â I want to draw a picture of this object's location over time.

Â And I know that, well, this pendulum is vibrating back and forth.

Â And if it's got a period of 2 seconds and

Â it's a simple harmonic oscillator, we should be able to represent its

Â position over time as a sine or a cosine curve, something like this.

Â Where I can tell that the object in motion has repeated itself in a full 2 seconds.

Â And is not done so, then let's say one second.

Â I'll also know that it's giving me a little bit of information

Â about the amplitude, so

Â this will be my positive amplitude, this will be my negative amplitude.

Â In this case, something like .3 of a meter.

Â .3 of a meter.

Â And so taking a look at this, it asks you for two points.

Â Where is it located at .75 seconds?

Â So somewhere get about here.

Â Where is that location?

Â On the scratch it would be, where is it here on the Y axis?

Â And that 1.5 seconds here.

Â Well automatically I have my answer for 1.5 seconds.

Â I can tell by the way I've drawn that when it oscillates it's going to be

Â at the amplitude, which it doesn't say in this problem,

Â assuming that it went to the positive direction first and

Â reached this first positive amplitude here at .5 milliseconds,

Â that means it ends at a negative .3 meters at 1.5 seconds.

Â So I've already got the second portion of this question.

Â At 1.5 seconds, the mass is located

Â at a location of 0.3 meters amplitude.

Â And if I wanted to give a direction with some of the assumptions I made,

Â I'd said that'd be negative 0.3 meters.

Â Okay, it's a little more difficult to say precisely where this

Â thing is located at this 0.75 seconds location.

Â So to do that I'm going to use our sine or our cosine function.

Â Notice that our wave started at equilibrium, and

Â I did that very specifically since I want to know where this thing is located after

Â passing through equilibrium, that's why I started the motion at equilibrium.

Â So that when I am one second later or

Â two second later, it actually answers the question being asked.

Â So since this graph starts at zero, that is a prime time to use your sine version

Â of this expression to tell us where the mass located at any point in time.

Â Remember that at time 0, when t = 0,

Â sin will = 0 and therefore x will = 0.

Â That's why sin fits the behavior that we're analyzing here.

Â Let's plug in the numbers that they're giving us.

Â Remember that a frequency is the same thing as 1 over the period, and

Â we know that that's 2 seconds.

Â So my equation will look something like

Â this: 0.3 sin (2

Â pi/2(t)) is my function.

Â Two will cancel out, and now if I want

Â to know where is this object located at any point in time, I can sub in.

Â So x equals 0.3 sin of pi times 0.75 seconds.

Â Make sure your calculator is in radian mode,

Â otherwise you won't get the correct answer.

Â I get 0.212 meters, which is less than the amplitude,

Â which is a good check, we haven't made some egregious mistake.

Â It's also important to point out, you could also check our answer from earlier.

Â By subbing in instead of .75, 1.5 seconds, and

Â I promise that also turns out to give you the -0.3 liters.

Â It's important to remember, in the absence of friction,

Â springs are a conservative force.

Â That means energy is conserved.

Â And we can use our information about energy to solve

Â lots of interesting problems using springs and simple harmonic oscillators.

Â This one's a good example.

Â It starts off with a mass attached to a horizontal spring on a frictionless

Â surface, and for all of these problems, we assume massless springs.

Â Now that mass is then stretched two centimeters from equilibrium and

Â released, so it's stretched, perhaps, a little further out this way.

Â That means there will be a force to the left on this mass.

Â It will have gained some speed when it reaches back to the equilibrium position,

Â it will come back to the other side over here, and

Â now I have a force to the right on it.

Â And it will oscillate back and forth between this amplitude.

Â In this case, the original stretch,

Â because it starts from rest, which is 0.02 of a meter.

Â Now let's see what it asks for.

Â The first thing that it wants to know,

Â determine the maximum speed of the mass and where it occurs.

Â Looking at my patterns,

Â I know that the mass land amplitude has all spring potential energy.

Â That's where it stops to turn around but

Â at equilibrium, that's where it has its fastest speed.

Â That's where it's all kinetic energy.

Â No spring potential energy.

Â So, to solve that problem for finding the actual speed,

Â I'm going to use conservation of energy.

Â The sum of energy before,

Â at some point initial, will equal the sum of energy at some point final.

Â In this case, we have an initial spring potential, and

Â then we'll have a final kinetic translational energy.

Â However you'd like to represent that.

Â Remember subscripts are just for you.

Â Whatever you find helpful.

Â So 1/2kx squared is the initial spring potential energy.

Â 1/2mv squared will give us the final

Â linear translational energy of the sliding block.

Â 1/2 on both sides will cancel out, and I can sub in some of my numbers here.

Â 40 for k. It told me it was stretched 2 centimeters,

Â so 0.02 of a meter.

Â Don't forget to square your term.

Â Now this is a pretty large block,

Â 20 kilograms, so I'd expect a fairly small velocity.

Â And that's what I end up getting when I solve.

Â A velocity here of 0.028 meters per second.

Â Now that's the max speed and where does that occur?

Â So my answer as to to where that will occur, and that will be at

Â the equilibrium position.

Â So that's part A of this problem.

Â Let's see what part B asks.

Â What is the speed of the mass at one centimeter from equilibrium?

Â Now, looking at my diagram, that might be, let's say, this location here.

Â Where we're halfway.

Â Now, keep in mind the spring is still compressed.

Â It's not at its equilibrium position, so it has some speed,

Â some kinetic energy, as well as some spring potential energy.

Â So we need to keep both in mind.

Â So let's go ahead and take a look at that.

Â That means when I get to that location,

Â let's say energy before equals energy after.

Â Then, again,

Â I will have all string potential at the beginning, 1/2 k x squared.

Â Remember, you can pick any points, so at this point, if you wanted to, instead,

Â select your initial point as the equilibrium position,

Â where it has all kinetic energy, that would be great.

Â 1/2 mv squared + 1/2 kx squared.

Â It's going to have some potential for

Â the spring, and some kinetic for its motion.

Â Again, 1/2 cancel out.

Â I can sub in numbers again.

Â 40 for the spring constant It was initially 0.02 of a meter

Â from equilibrium.

Â Remember, x doesn't represent the total length of the spring,

Â it's how far the spring has been stretched.

Â The mass again is 20 kilograms, our base unit for that.

Â V squared is what I'm looking for.

Â Again, our k is 40.

Â Except now I'm at a point where I'm only stretched one centimeter from equilibrium.

Â Notice that this term here on the right that wasn't here before

Â will have to be subtracted over to the left, and

Â will give us a smaller numerator, a smaller v in our answer.

Â Our answer better be something less than the maximum.

Â We've all ready solved what the maximum speed is.

Â So if you come up with something larger, you know you've made a mistake.

Â I get something smaller, that's good, it's what I was looking for.

Â That's the speed.

Â The problem then goes on to ask a couple of other questions pretty quickly.

Â Now this was part B.

Â Let's see what part C says.

Â Determine the maximum restoring force and where it occurs.

Â Now I know that the spring has no restoring force when it's at equilibrium.

Â That's what it means to be at equilibrium.

Â The spring is not compressed and therefore there is no force.

Â It would give the most where it is stretched completely.

Â In this case, at either amplitude would work.

Â And so if I want, for this portion,

Â reaching covers again, to calculate the max restoring force and

Â where that occurs, I'm going to use our equation for the force of a spring.

Â Notice, I normally drop the negative sign in this equation,

Â you will see it written with a negative sign in many instances.

Â The negative signs of forces helps you to keep track of direction.

Â I'm just looking for the magnitude here.

Â So I'm going to go ahead and drop it.

Â My spring constant was 40.

Â I know the most it gets stretched in this scenario is at the amplitude.

Â That will give me the largest force.

Â And so I have a force of 0.8 Newtons,

Â and it occurs at the amplitude.

Â It then also asks for our last portion because this is portion C.

Â Part D says determine the max acceleration and where it occurs.

Â Well, because of Newton's Second Law, the sum of forces equals MA.

Â Where you have the max force, you have your max acceleration.

Â So I automatically know the max acceleration will occur at

Â the amplitudes, as well.

Â I just need to calculate what is that acceleration.

Â So for the last portion, Newton's Second Law, sum of forces equals MA.

Â I know there's a small force.

Â Okay, the force on this, .8.

Â I know that there's a small force on this object, the mass itself.

Â 20 kilograms, fairly large times A.

Â Solving for the mass acceleration then, it's going to be a small value,

Â which is what yielded the small velocity.

Â Make sure to give units.

Â Acceleration, 0.04 meters per second squared, and

Â it also occurs at the amplitude.

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