I’ll need more information about what you don’t understand to give a satisfying answer, the discussion is above so pointing specifically at the bit you don’t get is useful. The mass also bounces higher than x=0. But anyway; the force/distance curve for x0. Imagine hanging a mass off a spring, and pulling down on it. Then, the spring will shoot up and the mass will do a little hop in the air before falling again. This is exactly the scenario in this question. Hopefully this gives you a little more intuition?
Hi, thank you for the wonderful web. Question: u claimed that the ball has 100 J energy but that assumes the ball rest on x=0 eventually. However, it should rest on a point with no force, which is x=-10. If It stays at x=0, the force should move it into the left half plane and keeps moving. I believe the energy used should be 10×10 (x=0 to 10) + 10×10/2 (x=-10 to 0) so the distance should be 150 m.
Hi, I would appreciate it if you could expound on the “Max KE = 10 hence speed = 2xsq5) for part b1. Since the max KE was found to be 150 in part a2, and it dissipates 100 joules of energy due to friction, should it not have 150 -100 = 50 joules for max KE after it starts oscillating between x= – 20 and x = 0 ?Thank you
For the last part of the question, do you mind explaining it a bit more? I’m a bit unclear about the complex method as well as why we would start with F=10 rather than 9 in the simple method. I’m just hoping you haven’t slept yet over
I dont understand how you found the area under the curve to be 100, wouldnt it be the area of the big triangle minus the area of the small triangle? If so, i got that value to be 150.
Can you please explain this point,
Thanks in advance.
Have you studied how SHM is governed by second order differential equations like this? If not then this will look like magic, but it’s not really that hard. I recommend this site: http://www.animations.physics.unsw.edu.au/jw/DifferentialEquations.htm#SHM, which illustrates some of the ideas with animations (don’t bother with anything after the damped and forced oscillations unless you are particularly interested).
No I haven’t. Thank you for the link. There is just one bit I don’t really understand: ‘We can solve these in terms of A and φ, first by dividing the two equations, then by squaring them and adding. So, for these given initial conditions, we can find a combination of the constants A and φ, so this is the general solution.’ I don’t know if I’m just being stupid but it doesn’t seem very clear how they want you to divide etc… (what by what for example).
I get they’re trying to show you how to isolate A and phi to find them in different situations (or not?) like here, but it isn’t clear to me exactly how. Sorry for so many questions!
You’re right — it’s just trying to express A and phi in terms of x0 and v0, but it’s a bit cryptic. What it means by ‘dividing the two’ is ‘divide the first by the second to get x₀/v₀ = (tan φ)/ω’, so that we derive that φ is the inverse tan of ωx₀/v₀. What it means by ‘squaring and adding’ is ‘multiply the first by ω and square it, and add to the square of the second to get ω²x₀² + v₀² = ω²A²’ so that we derive that A is √(x₀² + v₀²/ω²).
It’s a pleasure. And thanks for your questions: the more Q&As we have on the site, the better the explanations get for everybody, so by asking questions you’re doing everyone a favour.
Could you be more specific please? In this question? Or in general for SHM. If it’s in the general then any notes on SHM should help you out.
We have that (d/dt)^2 x = – k/m x = – w^2 x. This gives w = sqrt(k/m). SHM is I believe in the A-level syllabus; it is the most important second order differential equation.
Sorry if I didn’t make it clear. Using the link above I understand that x= Asin(wt +phi) and from the question (d/dt)^2(t)= -x-10. And the SHM starts at t = sqrt(2) where x=0 and v = 10sqrt(2) so using these value I tried to solve for A and phi and got to phi= -wsqrt(2) and A = sqrt(200/w^2) however I don’t know how to find w and get to your answer of A=10sqrt(3).
On part a3, why do you treat the acceleration like simple harmonic motion instead of using F = ma and m = 1 to say F = a, which would make the acceleration graph the same as the force graph?
(I suppose the graph for velocity would then be the same as kinetic energy. But, a = v(dv/dx) if I understand M3 kinematics, so integral of a dx would be 0.5v^2 — which is the same as the energy graph, since KE = 0.5mv^2 and m = 1, so KE =0.5v^2 = a … I don’t know if that makes sense 😦 )
Thank you!
Your reasoning in the first paragraph is correct, but the graph that the question shows is force v distance, and the graph they are asking you to sketch is force v time. So you need to go from the differential equation d²x/dt² = -x to be able to find dx/dt and d²x/dt² in terms of t. That’s when it helps to spot that this is just SHM.
How rigorous do they expect answers to be in general? I didn’t notice the SHM property of this question and consequently got the shock of my life when I saw this..
I wouldn’t worry too much about this question — it is by far the hardest question that they’ve ever asked. It took me a couple of hours to get the answer out properly. If you look at the examiners’ report for that year you’ll notice that nobody got more than 83% on the paper as a whole — my guess is that nobody got more than about 10/20 on this question.
Having said that, though, the more rigorous your answers the better. The one thing the examiners are looking for is that you can turn vague words into accurate mathematical content (e.g. equations, diagrams), and then do something useful with it. Every line of maths that you can get down (as long as it’s right) is worth something.
Hi please could you explain your first line of working for a1. The paticular bit where you say Ek = fFdx = 100 I dont understand that line but the other lines i do
The graph is really two graphs, one for x > 0 and one for x < 0, with a discontinuity at x = 0. Ignore the x > 0 part and you can see that it does look like SHM when x < 0. Specifically, when x < 0, the equation of the graph is F = -(x + 10), which looks like a mass/spring equation.
Try working out the graph of force against height for a 1kg mass on a spring that has spring constant 1N/m, given these conditions: the acceleration due to gravity g is 10m/s/s; when the height is positive the spring is slack; when the height is zero the spring is just about to start extending; and the more negative the height the more extended the spring gets. You should find that you get the same graph that you see in the question.
I’m really sorry to be a pain, but I cannot see how you generated the equation relating xi+1 and xi, and also where the 9/11 has come from. I presume it relates to the force acting?
When the mass is stationary at xi, the PE is 10xi, and the KE is zero.
The mass then moves a distance xi (with friction acting) and enters the x<0 zone. Some time later is comes bouncing back from the x<0 zone, travels a further distance xi+1 (with friction acting), and ends up stationary again at xi+1.
When the mass is stationary again at xi+1, the PE is 10xi+1 and the KE is zero.
In moving from the stationary point xi to the stationary point xi+1, the mass did work in overcoming friction by travelling xi (from the first stationary point to the x=0 point), and then travelling xi+1 (from the x=0 point to the second stationary point). So, by conservation of energy:
10 xi – xi – x(i+1) = 10 x(i+1), which means that 9xi = 11xi+1, so xi+1 = (9/11) xi
OK, I’ll describe each of the bullet points in turn:
When x<0 there is no friction, so the mass doesn’t lose any energy. So if it crosses the x=0 boundary on the way to the x<0 zone with velocity -v, it will eventually come bouncing back across the x=0 boundary with velocity +v. And also it is worth noting that this means the section of the graph for x<0 is symmetric about the x axis.
In the x>0 zone, the mass is stationary at the point of maximum x, it then accelerates back to the x0 zone, where is decelerates to a new point of maximum x. But because the mass loses energy to friction each maximum x point has a smaller value of x than before. We write the initial maximum as x0, and sebsequent ones as x1, x2, … xi
When the mass is at +x it has PE of 10x, and when it moves against friction a distance x it loses energy x. So we can generate an equation relating xi+1 to xi, and when we plug in the numbers above we get xi+1 = (9/11) xi
After a long time the maximum x value tends to zero, so ultimately we just end up with the mass bouncing up and down on the string endlessly.
Having considered the KE it should be easy to work out what this implies for v, which is proportional to the square root of KE. Just be careful to the the sign of v right — it starts negative.
Finally in part b2, you should be able to see that the mass only disspates energy when x>=0 and it does this by moving against a frictional force of 1N. It starts with energy of 100J, which all has to be dissipated, so it must move 100m in the x>=0 zone.
Just for completeness, I have also summed up the infinite series to demonstrate that you get the same result. But this isn’t necessary in order to answer the question.
Oxford PAT 
Hi, could you explain how does the mass end up bouncing between -20 and 0?
I’ll need more information about what you don’t understand to give a satisfying answer, the discussion is above so pointing specifically at the bit you don’t get is useful. The mass also bounces higher than x=0. But anyway; the force/distance curve for x0. Imagine hanging a mass off a spring, and pulling down on it. Then, the spring will shoot up and the mass will do a little hop in the air before falling again. This is exactly the scenario in this question. Hopefully this gives you a little more intuition?
Hi, thank you for the wonderful web. Question: u claimed that the ball has 100 J energy but that assumes the ball rest on x=0 eventually. However, it should rest on a point with no force, which is x=-10. If It stays at x=0, the force should move it into the left half plane and keeps moving. I believe the energy used should be 10×10 (x=0 to 10) + 10×10/2 (x=-10 to 0) so the distance should be 150 m.
The question asks how much distance is covered in the region where x >= 0, so where it comes to rest is not important.
Hi, I would appreciate it if you could expound on the “Max KE = 10 hence speed = 2xsq5) for part b1. Since the max KE was found to be 150 in part a2, and it dissipates 100 joules of energy due to friction, should it not have 150 -100 = 50 joules for max KE after it starts oscillating between x= – 20 and x = 0 ?Thank you
For the last part of the question, do you mind explaining it a bit more? I’m a bit unclear about the complex method as well as why we would start with F=10 rather than 9 in the simple method. I’m just hoping you haven’t slept yet over
I dont understand how you found the area under the curve to be 100, wouldnt it be the area of the big triangle minus the area of the small triangle? If so, i got that value to be 150.
Can you please explain this point,
Thanks in advance.
Never mind, I didnt read the bit where it says it started of at x=10 haha.
How did you go from d^2x/dt^2 = -(x+10) to working out that it is in SHM with ‘x = 10root3(t+phi)? I’m really confused…
This is in part a3 sorry
Have you studied how SHM is governed by second order differential equations like this? If not then this will look like magic, but it’s not really that hard. I recommend this site: http://www.animations.physics.unsw.edu.au/jw/DifferentialEquations.htm#SHM, which illustrates some of the ideas with animations (don’t bother with anything after the damped and forced oscillations unless you are particularly interested).
No I haven’t. Thank you for the link. There is just one bit I don’t really understand: ‘We can solve these in terms of A and φ, first by dividing the two equations, then by squaring them and adding. So, for these given initial conditions, we can find a combination of the constants A and φ, so this is the general solution.’ I don’t know if I’m just being stupid but it doesn’t seem very clear how they want you to divide etc… (what by what for example).
I get they’re trying to show you how to isolate A and phi to find them in different situations (or not?) like here, but it isn’t clear to me exactly how. Sorry for so many questions!
You’re right — it’s just trying to express A and phi in terms of x0 and v0, but it’s a bit cryptic. What it means by ‘dividing the two’ is ‘divide the first by the second to get x₀/v₀ = (tan φ)/ω’, so that we derive that φ is the inverse tan of ωx₀/v₀. What it means by ‘squaring and adding’ is ‘multiply the first by ω and square it, and add to the square of the second to get ω²x₀² + v₀² = ω²A²’ so that we derive that A is √(x₀² + v₀²/ω²).
Thank you so much for your help. Your responses have really been excellent.
It’s a pleasure. And thanks for your questions: the more Q&As we have on the site, the better the explanations get for everybody, so by asking questions you’re doing everyone a favour.
How do you find omega
Could you be more specific please? In this question? Or in general for SHM. If it’s in the general then any notes on SHM should help you out.
We have that (d/dt)^2 x = – k/m x = – w^2 x. This gives w = sqrt(k/m). SHM is I believe in the A-level syllabus; it is the most important second order differential equation.
Sorry if I didn’t make it clear. Using the link above I understand that x= Asin(wt +phi) and from the question (d/dt)^2(t)= -x-10. And the SHM starts at t = sqrt(2) where x=0 and v = 10sqrt(2) so using these value I tried to solve for A and phi and got to phi= -wsqrt(2) and A = sqrt(200/w^2) however I don’t know how to find w and get to your answer of A=10sqrt(3).
On part a3, why do you treat the acceleration like simple harmonic motion instead of using F = ma and m = 1 to say F = a, which would make the acceleration graph the same as the force graph?
(I suppose the graph for velocity would then be the same as kinetic energy. But, a = v(dv/dx) if I understand M3 kinematics, so integral of a dx would be 0.5v^2 — which is the same as the energy graph, since KE = 0.5mv^2 and m = 1, so KE =0.5v^2 = a … I don’t know if that makes sense 😦 )
Thank you!
Your reasoning in the first paragraph is correct, but the graph that the question shows is force v distance, and the graph they are asking you to sketch is force v time. So you need to go from the differential equation d²x/dt² = -x to be able to find dx/dt and d²x/dt² in terms of t. That’s when it helps to spot that this is just SHM.
Ohh, thank you! 🙂
How rigorous do they expect answers to be in general? I didn’t notice the SHM property of this question and consequently got the shock of my life when I saw this..
I wouldn’t worry too much about this question — it is by far the hardest question that they’ve ever asked. It took me a couple of hours to get the answer out properly. If you look at the examiners’ report for that year you’ll notice that nobody got more than 83% on the paper as a whole — my guess is that nobody got more than about 10/20 on this question.
Having said that, though, the more rigorous your answers the better. The one thing the examiners are looking for is that you can turn vague words into accurate mathematical content (e.g. equations, diagrams), and then do something useful with it. Every line of maths that you can get down (as long as it’s right) is worth something.
Hi please could you explain your first line of working for a1. The paticular bit where you say Ek = fFdx = 100 I dont understand that line but the other lines i do
The line says that the energy is the integral of force with respect to distance, i.e. the area between the force-distance curve and the x-axis.
Could you explain how you get SHM for part a3? The force vs. position graph doesn’t look anything like SHM, right? Thank you!!
Hi Francis, thanks for your question.
The graph is really two graphs, one for x > 0 and one for x < 0, with a discontinuity at x = 0. Ignore the x > 0 part and you can see that it does look like SHM when x < 0. Specifically, when x < 0, the equation of the graph is F = -(x + 10), which looks like a mass/spring equation.
Try working out the graph of force against height for a 1kg mass on a spring that has spring constant 1N/m, given these conditions: the acceleration due to gravity g is 10m/s/s; when the height is positive the spring is slack; when the height is zero the spring is just about to start extending; and the more negative the height the more extended the spring gets. You should find that you get the same graph that you see in the question.
I’m really sorry to be a pain, but I cannot see how you generated the equation relating xi+1 and xi, and also where the 9/11 has come from. I presume it relates to the force acting?
When the mass is stationary at xi, the PE is 10xi, and the KE is zero.
The mass then moves a distance xi (with friction acting) and enters the x<0 zone. Some time later is comes bouncing back from the x<0 zone, travels a further distance xi+1 (with friction acting), and ends up stationary again at xi+1.
When the mass is stationary again at xi+1, the PE is 10xi+1 and the KE is zero.
In moving from the stationary point xi to the stationary point xi+1, the mass did work in overcoming friction by travelling xi (from the first stationary point to the x=0 point), and then travelling xi+1 (from the x=0 point to the second stationary point). So, by conservation of energy:
10 xi – xi – x(i+1) = 10 x(i+1), which means that 9xi = 11xi+1, so xi+1 = (9/11) xi
Please could you explain the (Xi)notation for the CoE part onwards? Thanks for all your help!
Or if you could, just to clarify I understand, a step-by-step explanation of part b would be great. Thank you!
OK, I’ll describe each of the bullet points in turn:
When the mass is at +x it has PE of 10x, and when it moves against friction a distance x it loses energy x. So we can generate an equation relating xi+1 to xi, and when we plug in the numbers above we get xi+1 = (9/11) xi
Finally in part b2, you should be able to see that the mass only disspates energy when x>=0 and it does this by moving against a frictional force of 1N. It starts with energy of 100J, which all has to be dissipated, so it must move 100m in the x>=0 zone.
Just for completeness, I have also summed up the infinite series to demonstrate that you get the same result. But this isn’t necessary in order to answer the question.