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# AP Physics 1 review of Waves and Harmonic motion

## Video transcript

Hookes law tells you how to find the force exerted by an ideal or linear spring and it's a simple law it tells you the amount of force that spring is going to exert will be proportional to the amount that spring has been stretched or compressed from its equilibrium or natural length which in equation form just says that the magnitude of the spring force is going to equal the spring constant multiplied by the amount the spring has been stretched or compressed note this X is not the length of the spring the X is how much that spring has been stretched from or compressed from the equilibrium position or the unstretched position so what's an example problem involving Hookes law look like let's say an ideal spring is hanging from the ceiling at rest and it has an unstretched length l-1 and then you hang a mass m from the spring at rest and it stretches the spring to a length L - what is an expression for the spring constant of the spring so the force of gravity has to be balanced by the spring force that means the magnitude of the spring force is equal to the magnitude of the force of gravity the spring force is always K times X what is X going to represent it's not going to be l1 or l2 X is how much the spring has been stretched from its equilibrium position which would be l2 minus l1 and if we solve for K we get mg over l2 minus l1 what's a simple harmonic oscillator a simple harmonic oscillator is any variable whose change can be described by a sine or cosine function what does that function look like it looks like this so the variable that's changing as a function of time which could be the vertical position of a mass on a spring the angle of a pendulum or any other simple harmonic oscillator is going to be equal to the amplitude of the motion which is the maximum displacement from equilibrium times either sine or cosine of 2 pi times the frequency of the motion times the variable T which since frequency is 1 over the period you could write as 2 pi over the period times the time variable T how do you know whether you use sine or cosine well sine starts at 0 and goes up and cosine starts at a maximum and goes down so if you know the behavior of your oscillator at T equals 0 you can decide whether to use cosine or sine something that's important to know is how to find the period of an oscillation the period of a mass on a spring is going to be 2 pi times the square root the mass connected to the spring divided by the spring constant K note that this does not depend on the amplitude if you stretch that mass farther it'll go faster and it has farther to go which cancels out and the period remains the same and the formula for the period of a pendulum which is a mass swinging on a string is going to be 2pi times the square root the length of the string divided by the magnitude of the acceleration due to gravity which also does not depend on the amplitude as long as the angles are small and it doesn't depend on the mass either how do you find this period on a graph well if you're given the graph of the motion of a simple harmonic oscillator as a function of time the interval between Peaks is going to represent the period or the time it takes for this oscillator to reset so it's an example problem involving simple harmonic motion look like let's say in a lab a mass M on earth can either be hung on a string of length L and allowed to swing back and forth with a period T pendulum or hung on a spring of spring constant K and allowed to oscillate up and down with a period T spring if a 2m mass were used instead of the one M mass what would happen to the period of the two motions well the period of a pendulum doesn't depend on the mass and so the period of the pendulum would not change the answer would have to be D what are waves waves or disturbances that travel through a medium and transfer energy and momentum over significant distances without transmitting any mass itself over those distances what is medium mean this is a fancy word for the material through which the wave can travel so you can classify a wave by the medium it's in but you could also classify waves by the type of disturbance you've created for transverse waves the disturbance of the medium is perpendicular to the wave velocity by wave velocity we mean the direction in which the disturbance travels and by oscillation of the medium we mean the direction in which the particles of the medium actually move for a wave on a string the particles move up and down but the disturbance travels to the right so this would be a transverse wave for longitudinal waves the oscillation or disturbance of the medium is parallel to the wave velocity the classic longitudinal wave is sound if a sound wave we're traveling rightward through the air it would look like a compressed region and the air itself would move back and forth right and left parallel to the direction the wave disturbances traveling which makes sound waves longitudinal for every type of wave the speed of that wave disturbance is going to be equal to the wave length of the wave divided by the period in other words if you watched a wave crest that wave crest would move one wavelength every period and since the speed is distance per time the speed of the wave crest would be one wavelength per period you can find the wavelength on a graph of y versus X by finding the distance between crests and if you're wondering why this doesn't represent the period it's because this is a graph of the wave versus X versus the horizontal position not the time you could make a graph of y versus the time and what that would represent is the motion of a single point on the wave for all moments in time and for this graph versus time the interval between Peaks is the period so if you get a graph of a wave you've got to check whether it's versus X or versus T if it's versus X peak to peak is the wavelength and if it's versus T peak to peak is the period and since 1 over the period is equal to the frequency we can rewrite the speed formula as the speed of a wave equals the wavelength of the wave times the frequency and the way it's given on the formula sheet on the AP exam is that the wave length of a wave is equal to the speed of the wave divided by the frequency now this formula confuses a lot of people though because they think if you increase the frequency that'll increase the speed of the wave but that's not true increasing the frequency will cause the wave length to decrease and the speed of the wave will remain constant the only way to change the speed of a wave is to change the properties of the medium itself in other words the only way to change the speed of the waves on water would be to change something about the water itself its density its salinity the temperature of that water changing frequency isn't going to change the speed of the wave and neither will changing the amplitude the only thing that changes the speed of the wave is changes to the medium itself and so what's an example problem involving waves look like let's say a sound lab is being conducted lab room with total cubic volume V and temperature T a speaker in the room is hooked up to a function generator and it plays a note with frequency F and amplitude a which of the following would change the speed of the sound waves increasing the frequency would just make that sound a higher note but it wouldn't change the speed of the sound wave increasing the temperature of the room is a change to the medium itself so this would change the speed of sound decreasing the amplitude is just going to make that sound seem softer and not appear as loud and decreasing the total volume of the space in the lab room doesn't actually affect the medium it just gives you less of the medium so the best answer here would be be the Doppler effect refers to the change in the perceived frequency when a speaker or a wave source moves relative to the observer if a wave source and observer are moving toward each other the wavelength of the sound wave is going to decrease according to that observer which would make the perceived frequency increase this happens since if the source is heading toward you as that speaker emits wave pulses the speaker moves toward the wave pulse it just emitted and on this leading edge the crests of the wave will be closer together since they're closer together the wavelength is smaller and the rate at which these crests are going to hit the observer is going to be higher so this observer is going to hear a higher frequency than is actually being played by the wave source when it is at rest and for the observer on the trailing edge since the wave source is moving away from the pulses it just sent in this direction these wave crests are going to be spaced further apart which increases the wave length and decreases the rate at which these crests are going to hit the observer so this observer is going to hear a frequency that's less than the actual frequency being played by the source when it's at rest so it's an example problem involving the Doppler effect look like let's say the driver of a car sees that they're heading straight toward a person standing still in the crosswalk so the driver continuously honks their horn and emits a sound of frequency F horn this is the frequency the horn plays when the car would be at rest and the driver of the car also simultaneously slams on the brakes and skids to a stop right in front of the person standing in the crosswalk what would that person in the crosswalk hear as the car is skidding to a stop well since that car is heading toward the person those wave crests are going to be spaced closer together so this person will hear a smaller wavelength and a higher frequency but as the car slows down this effect becomes less and less dramatic and once the car stops the wave crests will be spaced out their normal spacing and the person will just hear the regular horn frequency of the car so first this person is going to hear a higher frequency but that will eventually just become the actual frequency of the horn once the car stops and there's no more relative motion between the person and the car when two waves overlap in the same medium we call it wave interference or wave superposition while those waves are overlapping they'll combine to form a wave shape that will be the sum of the two waves in other words while the two waves are overlapping to find the value of the total wave you just add up the values of the individual waves so if you overlap two waves that look identical they would combine to form a wave that's twice as big and we call this constructive interference and if you overlap two waves that are 180 degrees out of phase they'll combine to form no wave at all we call this destructive interference so even though while the waves are overlapping you add up their individual values to get the total wave after these waves are done overlapping they'll pass right through each other so if I sent a wave pulse down a string at another wave pulse when these two pulses overlap the string would be flat but shortly after that the wave pulses would continue on their way unaffected they don't bounce off of each other or cause permanent damage it's only while they're overlapping that you'll get the wave interference so it's an example problem involving wave interference look like let's say two wave pulses on a string head toward each other as seen in this diagram to the right and we want to know what would be the shape of the wave when the wave pulses overlap so to find the total wave we'll add up the values of each individual wave the blue wave is going to move to the right and the red wave is going to move to the left and we'll add up the individual values zero of the red wave plus negative two units of the blue wave will add up to negative two units for the total wave and then positive two units of the red wave plus negative two units of the blue wave is going to equal Oh units for the total wave again positive two units of the red wave plus negative two units of the blue wave add up to zero units for the total wave and then zero units for the red wave plus negative two units for the blue wave is going to equal negative two units for the total wave so our total wave will look like this since this pyramid just took a bite out of this blue rectangular wave how do you deal with standing waves on strings well to get a standing wave at all you need waves overlapping that are going in opposite directions but even if you have that you're not necessarily going to get a standing wave only certain allowed wave lengths will create a standing wave in that medium and what determines the allowed wave lengths is the length of the medium and the boundaries of that medium in other words the ends of a string could be fixed or loose if the end of the string is fixed it's going to be a displacement node note is the word we use to refer to points that have no displacement and if the end of a string is loose that end would act as a displacement antinode antinode being a point that has maximum displacement what would these standing waves look like standing waves no longer appear to move along the medium they just oscillate back and forth in place so this anti node would move from the top to the bottom back to the top but you wouldn't see this crest move right or left hence the name standing wave however the nodes stay put there's never any displacement at a node once you've determined the boundary conditions and the length of that medium the possible wave lengths are set because the only allowed standing waves have to start at a node and end at a node the fundamental standing wave refers to the largest possible wave length standing wave and in this case it would be one half of a wave length so the length of the medium would have to equal one half of the wavelength of the wave similarly for the second harmonic we still have to start and end at a node so the next possibility would be an entire wave length which means the length of the medium would equal one wavelength of the wave and if you get to the third harmonic this is three halves of a wavelength and there's no limit you can keep going here to excite these higher standing waves you've got to keep increasing the frequency because you'll keep decreasing the wave length of the distance between Peaks so what would a standing wave look like if one of the ends were loose in that case that end would be an anti-node and the fundamental standing wave would only take the shape of a fourth of a wavelength since it has to go from a node to an anti-node that means the length of the string would have to equal one-fourth of a wavelength the next possible standing wave would be 3/4 civil wavelength and the next possibility would be 5/4 of a wavelength and again this progression keeps on going so what's an example problem involving standing waves on strings look like plus a 1 end of a string of length L is attached to the wall and the other end is fixed to a vibrating rod a student finds that the string sets up a standing wave as seen here when the frequency of the rod is set to F naught what are the speed of the waves on the string well we know the string length is L and we can figure out how much of a wavelength this is from here to here would be one wavelength and there's another half so the string length L is equal in three halves of a wavelength or in other words the wavelength here is 2l over three and we know the speed of the wave is always wavelength times frequency so the speed of the wave here is going to be 2l over 3 which is the wavelength of the wave times the frequency and so the best answer is D how do you deal with standing waves in tubes so just like standing waves in strings the wavelength of the standing waves in a tube are determined by the length of the tube and the boundary conditions of that tube but this time instead of the string shaking you're creating a standing wave out of sound waves now for the boundary conditions of this tube an open end is going to act like a displacement anti-node since the air at an open end can oscillate wildly and you get maximum air disturbance but a closed end of a tube is going to act like a displacement node since there will be no air disturbance at a closed end so what if both ends of the tube were open giving us antinode antinode standing waves well the first possibility of the largest wave length standing wave would go from anti node to anti node and this is one half of a wave length so the length of this tube would have to equal 1/2 of a wave length the next possibility would still go from anti node to anti node and this is equal to one whole wavelength it might not look like it but from Valley to Valley is a whole wavelength so the length of this tube would equal one wavelength and the next possibility would equal three halves of a wavelength and you should know this is the same progression that we had for node nodes strings so whether both ends are anti nodes or both ends our nodes if both boundary conditions are the same you get this same progression that goes half of a wavelength one whole wavelength three halves of a wavelength basically any integer or half integer wavelength what if we closed one of the ends of this tube if we closed one end of the tube that end would become a displacement node since the air can't move at that position which would make it a node and it would have to go to the open end which is an anti node so the largest possibility this time would be one-fourth of a wavelength the next possibility would still go from node to anti node and this would be three-fourths of a wavelength and if you notice this is exactly the same as when we had node anti node strings we had the same progression of lambda over four three lambda over four five lambda over four any odd integer lambda over four were the allowed standing wave lengths so if one end has a different boundary condition from the other end this is going to be the progression of allowed lengths of the medium so what would an example problem involving standing waves and tubes look like let's say you blow over the top of a tube that's open at both ends and it resonates with a frequency F knot if the bottom of the tube is then covered and air is again blown over the top of the tube what frequency would be heard relative to the frequency heard when both ends were open well when both ends are open we know the standing wave is going to be an anti node to anti node which is one half of a wavelength which would equal the length of that tube so the wavelength would be two times the length of the tube but when we close one of the ends we turn that end from an anti node into a node so we'd have to go from an anti node to a node which is only one-fourth of a wavelength so one-fourth of a wavelength would equal the length of the tube that means lambda equals 4l this wavelength doubled so what would that do to the frequency where we know V equals lambda F and we didn't change the media here so the speed is going to remain the same so if we double the wavelength we'd have to cut the frequency in half in order to maintain the same speed of the wave so when we close the bottom of this tube we'd hear half the frequency we heard when both ends were open beep frequency refers to the phenomenon where two waves overlap with different frequencies when this occurs the interference of the waves at a point in space turns from constructive to destructive back to constructive and so on which if this were a sound wave you'd perceive as a wobble in the loudness of the sound and the reason this happens is that if these waves started in phase and they were constructive since they have different frequencies one wave would start to become out of phase with the other eventually becoming destructive which would be soft but if you wait longer one of these Peaks catches up to the next peak in the progression and the waves again become constructive which would be loud again and the time this takes to go from loud to soft to loud again is called the deep period but more often you'll hear about the beat frequency which is just one over the beat period so the beat period is the time it takes to go from loud to soft back to loud and the beat frequency is the number of times it does that per second how do you determine the beat frequency or the beat period well the formula use to find the beat frequency is actually really simple you just take the difference of the frequencies of the two waves that are overlapping if there is no difference if these waves have the same frequency you'd have a beat frequency of zero which would mean you hear no wobbles at all the further apart these two frequencies get the more wobbles you would hear per second and then to find the beat period you could take one over the beat frequency so what an example problem involving beat frequency look like let's say these two waves were overlapping and we want to determine the beat frequency the period of the first wave is four seconds that means the frequency of the first wave is one over four or 0.25 Hertz and the period of the second wave is two seconds which means the frequency is 1 over 2 or 0.5 Hertz to get the beat frequency you subtract one frequency from the other 0.5 minus 0.25 would be 0.25 Hertz
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