Phys260 Gupta HW2 SOLUTIONS (Oscillations/Waves) REQUIRED READING

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1 Notes: So first half of this HW is still on ideas from oscillations. Problems 2 and 3 blends in ideas from oscillations with some ideas about collisions from last semester. Problems 4-6 touch upon basic ideas on waves. For Oscillations it is still Knight ( ) or Tipler ( ). For waves in this HW, Knight ( ) and Tipler ( ). 1. Consider a ball of mass 100 grams tied to a very light string from the ceiling. The length of the string is 1 meters. (This is just like a simple pendulum setup). I move the ball to the right to an angle 10 degrees and release it (at t=0), so it starts oscillating (see figure). (a) What is the speed of the ball 0.2 seconds after I release it? [This solution contributed by Eric Kuo] I know how to find the position of the ball as a function of time. From there I can find the velocity at any time. There are two ways to think about this problem: using forces or using torques. They both lead to the same answer, but in your homework you probably went one way or other. Feel free to check the method you did on your homework, but before you read the other solution, I suggest you try it for yourself! I think the solution will do more for you if it s helping you on something that you ve tried yourself. Thinking about FORCES: 10 degrees is a small enough angle that we can approximate the motion of the ball as a horizontal oscillation. There are lots of ways to think about why we can approximate this. Here s one: solve for the x and y component of the ball s position using sin and cos. Then, use the small angle approximations: sinθ θ and cos θ 1. What is this approximation telling us? That the height of the ball is almost constant (L) for all small angles, but the horizontal position is changing as the angle changes (x = L θ). This is why we can approximate this as horizontal oscillation. So if the ball is only accelerating in the x-direction, then the net force in the y-direction must be zero, or else the ball would be accelerating up and down. (Ok, so in real life, the ball is actually moving up and down a little, so there is a net force in the y- direction. This is really the approximate net force that goes with our approximation of horizontal oscillation). We know that the y-component of tension of the string must be equal to mg, since it has to cancel the weight of the ball out. Using the angle of the string, we can find the x-component of the tension with a handy triangle. So the net force is in the x-direction and it s equal to F x = mgθ =

2 -mgx/l. Question: why is the force negative? Well, let s say that the origin is where the angle of the pendulum is zero. When the position of the ball is to the right, the force on it is to the left and vice versa; that s what a restoring force does. Mathematically, this means the sign of x and F x must be opposites, so we need a minus sign to make this happen. [Alternatively, you could think about mgsin(θ) as the force along the direction of the ball s motion that is speeding it up as the ball moves. The other forces T-mgcosθ provide the centripetal force; Using the approximation for small angles the net restoring force redues to mgθ note that this force changes direction depending on whether the ball is to the right or left of the equilibrium position; The other component, always acts along the direction of tension cannot change the speed of the ball and so can t act as the restoring force.] Now we have the net force, so we do what we do in oscillations: use Newton s 2nd law to find the position as a function of time. We know that the answer has to be cos or sin with some constants, so we just want to figure out what those constants are. We re thinking about it has a horizontal oscillation, so we want to write everything in terms of x, and not θ. The constant A is the amplitude of the oscillation. We know that the amplitude is the value of x when θ= θ 0 = 10 degrees rad ( A = x max = L θ 0 ). So x(t) = L θ 0 cos ( (g/l) t). The velocity of the ball is v(t) = dx/dt = -( (gl) θ 0 sin ( (g/l) t). To find v(0.2), now just plug in all the numbers in the problem (g=10 m/s 2, L = 1m, t = 0.2 s, θ 0 = rad). The speed after 0.2 seconds is m/s (which is positive, because the negative sign just tells you the direction of the velocity). Thinking about TORQUES: The torque on the ball is τ = -rfsinθ = -(L)(mg)sinθ and since 10 degrees is small enough for the small angle approximation, τ = -Lmgθ. The torque is negative, because it always wants to rotate the

3 pendulum back to equilibrium. This means the torque will always have the opposite sign as the angle. The tension of the string doesn t cause a torque on the ball because the tension of the string is always pulling in towards the center of the arc of the pendulum. Torques are applied to objects when the forces on them cause them to rotate about a pivot point. Only gravity ever has a component in the direction that the ball is moving. Mathematically, you can see that the tension causes no torque because the angle between r and F tension is 180 degrees and sin(180 degrees) = 0. So we know how much torque is trying to rotate this pendulum. My motivation to find the torque is that for forces causing oscillations in straight lines we use Newton s 2nd law, but for rotations we want to use the rotational version of Newton s 2nd law, τ = Iα. The value of I ball = ml 2 (if you forget why this is, I suggest checking back to your 161 notes on rotational inertia). Now we do the usual thing that we do in oscillations: we plug everything into Newton s 2nd law. We know the answer has to be cos or sin with some constants, so we want to figure out those constants using the equation we get. The constant A is the amplitude of the oscillation. We know that the amplitude is the maximum value of θ. This is when θ = θ 0 = 10 degrees rad. So θ (t) = θ 0 cos ( (g/l) t). The angular velocity of the ball is ω(t) = dθ/dt = -( (g/l) θ 0 sin ( (g/l) t). We also know the relationship between velocity and angular velocity is v = Lω = -( (gl) θ 0 sin ( (g/l) t). Now, just plug in all the numbers in the problem (g=10 m/s 2, L = 1m, t = 0.2 s, θ 0 = rad). The speed after 0.2 seconds is m/s (which is positive, because the negative sign just tells you the direction of the velocity). NOTES: I want to point out two things: 1) I did the problem two ways and got the same answer. It s a good check not just to see if you re getting the right answer, but also to see if you really understand the ideas behind how we are dealing with oscillations and 2) I didn t plug in all the numbers until the end, and it s

4 not just because I like working with variables (although I do). In the end, I get an expression that tells me the velocity at any time for a pendulum at any mass with any length and initial displacement. (b) One student has the following solution: Well the angular frequency for the pendulum is just ω = (g/l) and so the speed of the ball must be v = rω. Here r = l and so v = l (g/l) = (gl). What is wrong about the student s reasoning? Even without going through this student s reasoning, I can already see that the velocity cannot be (gl). When a pendulum swings, it speeds up on the way down and slows down on the way up. This student is saying that the speed for the pendulum doesn t depend on time, since g and l are constant. So where is the mistake in this reasoning? The ω s in ω = (g/l) and v = rω do not represent the same physical idea. The first is the angular frequency of the pendulum, it is 2π divided by the period of the pendulum. The second is the instantaneous angular velocity, how fast an object is rotating at some particular time. If you want to know the speed of the pendulum at a particular time then it should be clear that you want the second, the instantaneous angular velocity. In fact, there is no way that the first ω, the angular frequency of the pendulum, can tell you the speed of the pendulum at a specific time. The angular frequency depends only on the period and the period is just the time it takes to complete one cycle of the pendulum. This isn t enough information to tell you how fast the pendulum is moving at a particular time. Here s an analogy: imagine your roommate leaves to go to class in the physics building and comes straight back 2 hours later. Even though you know the time it took for your roommate to leave and come back, you couldn t say anything about how fast he/she was walking at a particular time in the trip. 2. (These solutions adapted from Birkett and Elby) Cart 1, of mass 1.0 kg, rolls down the ramp shown here (height = 30 centimeters) and collides with cart 2, of mass 2.0 kg. The carts stick together and roll toward the spring, which has spring constant 200 N/m and equilibrium length 15.0 cm. Neglect friction, air resistance, and the rotational inertia of the wheels (if the wheels are tiny and light, the rotational inertia can be quite small compared to the inertia of the cart). (a) Taking rightward as positive, sketch rough, non-numerical graphs of position, velocity, and acceleration vs. time for cart 2. Start your graph when cart 1 reaches the bottom of the ramp, and end your graph when the stuck-together carts reach the bottom of the ramp after rebounding off the spring.

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6 Before cart 1 reaches cart 2. Cart 2 is just sitting there motionless! After Collision, but before reaching spring. Steady (constant-velocity) rolling, which means the position increases steadily, the velocity is constant (flat), and the acceleration is zero. Of course, there is a spike in acceleration right at the moment of collision. So the cart2 accelerates during the small time of the collision, after which both carts are travelling at the same speed. Note that during the collision, we can t tell for sure the profile of the acceleration, velocity, or position (not enough information for that) so I drew something on an average! Spring compresses. Just like a block on a spring, with the spring compressing from equilibrium. Spring decompresses. Again, just like a block on spring, starting from rest from its leftmost point. After leaving the spring, but before reaching ramp. Steady (constant-velocity) rolling in the negative direction. (b) Find the closest distance to the wall reached by cart 2. The carts are closest to the wall at the moment of maximum spring compression, which occurs when the carts momentarily come to rest while reversing direction. Intuitively, the carts speed upon reaching the spring controls how far the spring compresses. Energy conservation bears this out: As the spring compresses, the kinetic energy carried by the carts converts into elastic potential energy in the spring. And from that potential energy, we can find the corresponding compression distance. So, to make progress, we need to know the carts speed and hence, their kinetic energy upon first reaching the spring. Can we find that speed using energy conservation alone? Not quite. Sure, as cart 1 rolls down the ramp, its gravitational potential energy converts neatly into kinetic energy. But during the inelastic collision, some of that kinetic converts into internal, thermal energy; the carts heat up slightly. Since we don t know how much kinetic energy gets lost in this way, we can t use energy conservation to find the post-collision speed. But since (i) the force by cart 1 on cart2 is always equal and opposite to that on cart 2 by cart 1 and (ii) there are no other forces in the horizontal direction, the change in horizontal momentum of cart 1 is equal and opposite to the change in horizontal momentum of cart 2. And we can use that to find the speed at which the carts hit the spring. In summary, we ll (1) use energy conservation to find v1, the speed of cart 1 before the collision; (2) use momentum conservation to find v2, the speed of the stuck-together carts after the collision; and (3) use energy conservation to find x max, the maximum distance by which the spring compresses from equilibrium. Given x max, we can figure out cart 2 s closest distance to the wall. So for the cart 1 that rolls down the ramp, we get

7 m 1 gh = 1 2 m 1v 1 2, and hence, v 1 = 2.42 m/s. This is the velocity of the cart1 at the bottom of the ramp before it hit cart2. For the collision, since p before = p after, we get m 1 v 1 = (m 1 + m 2 )v 2, which we can solve for v 2 to get v 1 /3 = m/s. (Hmmm, another way to think: since the combined mass of carts is 3 times the mass of cart 1 alone, after the collision the combined speed is 1/3 the original speed of cart 1) Now, at maximum compression all the combined kinetic energy of the carts after collision is stored in the spring as potential energy: 1 ( 2 m 1 + m 2 )v 2 2 = 1 2 kx 2 max, and hence x max = meters = 9.9 centimeters. [Alternatively, you could think about the motion of the carts when they are in contact with the springs as part of a cycle of oscillation an oscillation where the speed at the equilibrium position is m/s. We know that the speed at the equilibrium position is just x max ω where x max is the amplitude of this oscillation and ω is the angular frequency... and setting that would give us x max = v 2 /ω which if you substitute the angular frequency of that oscillation would turn out to be the same as you got using energy!] Finally, since the spring compresses 9.9 cm, cart 2 must be (15-9.9) = 5.1 cm from the wall at the time of (c) maximum compression. How much time does cart 2 spend in contact with the spring? Hint: You can think of the time when the carts are in contact with the spring as part of a period of an oscillation as if the carts were glued to the spring. Think about what fraction of that period do the carts remain in contact with the spring. If the carts became attached to the spring, a complete oscillation would be 1 Equilibirum 2 Leftmost point 3 Equilibirum 4 Rightmost point Equilibirum. But here, the carts lose contact with spring after completing the second leg of the cycle: Equilibirum 1 Leftmost point 2 Equilibirum. Our point: The carts stay in contact with the spring for exactly half of a complete oscillation. Therefore, the time spent in contact with the spring is half a period. So, we need to solve for T and divide it in half. Intuitively, the period depends on the stiffness of the spring, since a higher k leads to quicker oscillations. T should also depend on the mass; common sense says that heavier carts lead to more lackadaisical, longer-lasting oscillations. When we define angular frequency as ω = k m = 200 N/m 3.0 kg = 8.16 s -1,

8 and the period of oscillation is inversely proportional to w. T = 2π ω. Here, the period works out to be 0.77 seconds. The carts stay in contact with the spring for half that time: t = T/2 = 0.38 seconds. (d) This experiment is repeated, but with weights piled on top of cart 2, to make it heavier. Compared to your part (c) answer, does cart 2 now spend more time, less time or the same time in contact with the spring? Answer without doing new calculations. More time in contact with spring! As discussed in part (c), the carts stay in contact with the spring for half a period. So, this question asks whether the new scenario increases or decreases the period of oscillation. Using equations from above, we can jump to the answer. Since ω = T = 2π ω = 2π k /m = 2π m k, k /m, we get The spring constant k is the same as before, but the carts mass is now bigger. So, the period is now bigger; the carts spend more time in contact with the spring. Before moving on, we must see if this answer agrees with common sense, reconciling where necessary. Does it make sense that increasing the mass of carts makes the oscillations more lackadaisical (longer period)? Absolutely. [Here is a more subtle question: Does the total distance that the spring compresses change in the case when cart2 weights more? If it does, would that not also affect the time that the carts stay in contact with the spring?] 3. (Still working with the same setup as the previous problem The original setup; NOT the one in part(d) with added weights) A tenth of a second after cart 2 first comes in contact with the spring, how far is it from the wall? While touching the spring, the carts feel a proportional restoring force. Make sure you can work this force out and with that the expression for the position of the carts when they are in contact with the spring; don t just reach for formulas. The relevant conclusion is that, if we set x = 0 at equilibrium, as drawn here, then the carts position while in contact with the spring varies sinusoidally in time, in accordance with x = Asin wt, x = Acos wt, or some variant of those functions. Can we use one of those function to solve for x at time t = 0.10 s? Well, the amplitude, A, is the maximum displacement from equilibrium. We figured it out while solving part (b); the maximum compression is A = 9.0 centimeters = meters. And while solving part (c), we found the angular frequency. It s w = 8.16 s 1. So, we can indeed use a function like x = Asin wt or x = Acos wt to solve for x at a known time. We just need to figure out exactly which of those sinusoidal functions describes the carts motion!

9 Let s set t = 0 when the carts first reach the spring. At that moment, the left end of cart 2 is at x = 0 (the spring s equilibrium), and the cart s position is about to increase. So, the sinusoidal function describing the carts position is the one that starts at 0 and immediately goes up. Only sine fits this description. So, while in contact with the spring, the carts position is given by x = Asinωt. Using the amplitude and angular frequency discussed above, we get x = Asinωt = (0.099 m) sin 8.16 s -1 [( )( 0.10 s) ] = m = 7.2 centimeters. That s not the answer to this question. It s the distance by which the carts have compressed the spring leftward from equilibrium Remember, in that equation we just used, the quantity x is the displacement from the equilibrium position. The right end of the cart is therefore 15.0 cm x = 15 cm 7.2 cm = 7.8 cm from the wall. 4. [These solutions adapted from Birkett and Elby](Birkett and Elby) Suppose you make waves by repeatedly dipping your finger into a bathtub of water. Suddenly, you double the rate at which you dip. (a) Does the velocity of the waves go up, go down, or stay the same? Explain As we talked about in class, increasing the frequency i.e., producing more waves per second does not make the waves go faster. Wave speed is controlled entirely by the tightness (tautness) and density of the medium through which the waves travel, in this case water. Tapping the water at a higher frequency does not change the tightness or heaviness of the water. So, the waves travel at the same speed as before. However... (b) What about the wavelength? Explain....The waves are now closer together. More precisely, the wavelength gets cut in half. Here s why, intuitively. Each dip of the finger creates a new wave-crest (and trough). By doubling the dip rate of the finger, you cut in half the period, i.e., the time between successive dips. Hence, you cut in half the time that a given that a given crest gets to travel away from the finger before the next crest is created. Since the crests travel at the same speed no matter quickly or slowly you dip, cutting in half the time between successive crests also cuts in half the distance between. Which is another way of saying the wavelength is halved. The basic wave kinematic equation v = λf when rewritten as λ = v/f, captures this inverse proportionality between frequency and wavelength; doubling one cuts the other in half. Since v doesn t depend on the dip rate, increasing f causes λ to decrease by the same proportion. (c) Most people have the raw intuition that faster dipping creates faster motion. Does your part (a) answer hopelessly conflict with that intuition, or can you refine that intuition to help you understand what s going on? Explain. Hint: It might help some of you to think of a tiny cork bobbing up and down in the waves.

10 The intuition, when refined, is perfectly correct. When you dip at higher frequency, the waves get closer together, and hence the cork bobs up and down more quickly (higher average speed). So, faster dipping does create faster motion, where faster motion refers to the transverse particle speed (the individual water molecules bobbing up and down) rather than the wave speed (the speed of proportional of the pulses along the surface of the water. This is a classic example of refining, rather than abandoning, our intuitions. 5. (Birkett and Elby) One end of a flexible rope is attached to a wall. By shaking the other end back and forth at a rate of 3 times per second, Janice creates the wave pulse shown here. It travels rightward towards the wall. The gray squares were painted by a mischievous student on a window-pane immediately behind the rope. Those squares have a side length of 0.10 meters. (a) What is the amplitude of these waves? (b) What is the velocity of these waves? (c) Janice now shakes the rope again, but at a higher frequency. Compared to the waves in part (a), do these new waves travel faster, slower, or at the same speed? 5. [Solutons contributed by Colleen] (a) The amplitude of these waves is 0.05 meters. Amplitude is the maximum displacement from the equilibrium, which can be measured using the squares. The full swing of the wave is one full square = 0.1m So the deviation from equilibrium is half of that. (b) Think about a single point on the window pane. 3 wave-crests pass that point per second. So if I knew the how much distance is there between the crests, I say that he distance travelled by the way in 1 second equals three times the distance between two successive crests which gives me the velocity of the wave. From the snapshot provided, I can tell that each wave is 0.2 meters long (there are two squares from one crest to another). So the wave s velocity as it passes is (3 s -1 )*(0.2 m) = 0.6 m/s Or, another way to think about it: We are told that Janice shakes the rope 3 times per second, so it takes 1/3 second for her to create 1 wave pulse. We see that Janice has created 2.5 pulses, which took (2.5 pulses)*(1/3 second per pulse) = 0.83 seconds for the first pulse to get to its current position, which is 0.5 meters from the start. Therefore, the velocity of the waves is 0.5 meters/0.83 seconds = 0.6 m/s (c) The waves travel at the same speed! While the frequency is larger, the wavelength is smaller by exactly the right amount so that the velocity is the same. In general, the speed of a wave through a medium depends only on the medium s properties, not on how the wave was produced. Here is an analogy that could help make sense of it: Think about cars going down the road at some speed.

11 Each car here corresponds to a crest on the wave. You could double the number of cars on the road, still going the same speed, but that means the cars would have to be closer together. This would double the frequency, or # of cars that pass per minute, without increasing the speed of the cars. It is very tempting to keep the shape of the pulse (wavelength) the same and increase the number of cycles with increase in frequency and that makes it seem that higher frequency means that the pulse has gone out further. But if you pay attention to the mechanism of generation of the wave, you notice that the shape of the pulse (wavelength) would change as Janice shakes the rope more frequently generating pulses that have lesser distances between crests, but more number of crests. It is also tempting to think about how fast a bit of rope is moving. If Janice shakes her hand more frequently, then the bits of rope must be moving up and down faster (which is correct). And this sense of fastness (of the bits of rope) can make one think that the wave must have also travelled faster. But there are two kinds of faster involved here: the vertical speed of the bits of rope that does increases with frequency, and the horizontal speed of the pulse itself which remains constant! 6. (Birkett and Elby) A heavy weight is hanging from a very light string attached to the ceiling. A worker flicks the string near the ceiling, creating a wave pulse that heads toward the weight. As the pulse heads downward: (a) Does its speed increase, decrease, or stay constant? Explain. [This one I wrote up the solution] The speed remains the same. The speed of the pulse depends on the tension in the string and the linear mass density of the string. Neither of which is changing in the case given. Since the string is very light compared to the block that is hanging, the tension in the string is relatively unchanged from top to bottom (the top part of the string is supporting about the same weight as the bottom part). And so the speed of the pulse v = (T/μ) where T is the tension and μ the linear mass density of the string, remains unchanged. (b) Does its amplitude increase, decrease, or stay constant? Explain. The amplitude of the pulse also remains constant. If I think about two points on a string a little bit apart, then a change in amplitude of the pulse would mean that the bit of string that is lower goes to a different maximum distance than the point a little bit above it. If the point at top moves to an amplitude of 5 cm, there is no way it can push the point below it to move to 6 cm, unless the tension changes or the bit of string below is lighter. Could the bit below move just 4 cm? If the bit below moves with an amplitude of 4 cm then something is preventing it from reaching the 5 cm mark. One option is the pulse loses energy as it goes down so that the point that gets the pulse later moves through a lower amplitude. But there is nothing in the problem that suggests that the pulse is losing energy as it moves down. And there is nothing else that would prevent the bit of string

12 below to move to the amplitude of 5 cm, or push it further to an amplitude more than 5 cm. In other words, the amplitude of the pulse does not change! In reality well, the rope would have some mass and that would change the situation How? Well, HW3 solutions would talk more about that. Also, as the pulse passes by parts of the rope are stretched for some bit of time, before they relax again and this process heats up the rope by a little bit, essentially causing the pulse to lose some energy, which would affect the amplitude of the pulse. But these could be very small effects for a rope of light material and a small amplitude pulse that does not stretch the rope as much. (c) A common mistake people make when reasoning about a wave pulse is to think of it (perhaps unconsciously) as an object, analogous to a ball. How would someone thinking in these objectbased terms answer part (a)? How could you explain why that reasoning is incorrect? When I drop a ball down, or throw it down it goes faster and faster as it goes down, because the gravitational force is pulling it down or it accelerates due to gravity. Someone thinking about the pulse as an object could reason that gravity would accelerate the pulse, making it move faster as it goes down. Alternatively, someone could think about the energy of the pulse. As the ball goes down, the work done by the gravitational force on the ball increases its kinetic energy (or you could say it loses potential energy and gains kinetic it s all the same really). Similarly, one could argue that as the pulse goes down, the pulse gains kinetic energy. The response is that as the pulse is moving down, not a single bit of real matter, stuff, is actually moving down. Each little bit of the rope moves horizontally, not vertically! So, neither acceleration due to gravity nor work done by gravitational force come into play here. True the pulse moves down; but the pulse is not stuff and the ideas of acceleration as well as kinetic energy from Newton s Laws apply to stuff that has mass! It would be weird if someone asked me how much does the pulse weigh. And if the pulse cannot weight anything, does it make sense to think about its acceleration due to a gravitational force on it or work done on the pulse by the gravitational force? So, the reasoning about the pulse gaining speed is incorrect because it attributes mass to the pulse the pulse itself is massless, even though bits of stuff do move back and forth as the pulse washes over them.