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Ball Dropped Onto Spring

https://isaacphysics.org/questions/drop_ball_spring?board=f3ee3e64-45ea-47e3-91fe-a5c5f58f4c0b&stage=a_level

I got the right answer to this question but am a little confused about the process.
I was following someone elses working: https://imgur.com/gallery/ball-dropped-onto-spring-isaac-physics-problem-of15gWt

I dont understand why creating more variables is NECESSARY to solve the problem, specifically with u = Asin(wt+phi) since I thought this formula was for a distance quantity but here its used for 'u' which is equal to acceleration?

Why cant x (the extension of the spring) itself be = Asin(wt+phi) - I tried using this calculating x_dot and x_double_dot with this and it ended up giving me a contradiction.

Any help greatly appreciated.
(edited 3 weeks ago)
Reply 1
Original post by mosaurlodon
https://isaacphysics.org/questions/drop_ball_spring?board=f3ee3e64-45ea-47e3-91fe-a5c5f58f4c0b&stage=a_level
I got the right answer to this question but am a little confused about the process.
I was following someone elses working: https://imgur.com/gallery/ball-dropped-onto-spring-isaac-physics-problem-of15gWt
I dont understand why creating more variables is NECESSARY to solve the problem, specifically with u = Asin(wt+phi) since I thought this formula was for a distance quantity but here its used for 'u' which is equal to acceleration?
Why cant x (the extension of the spring) itself be = Asin(wt+phi) - I tried using this calculating x_dot and x_double_dot with this and it ended up giving me a contradiction.
Any help greatly appreciated.

The video looks a bit clearer but similar to the worked solution. The shm equation you seem to referring to is the transformed shm z = x + mg/k
z = Asin(wt+phi)
so shm about the equilibrium position x=-mg/k or z=0. A and phi are determined so that the initial position is z=mg/k and the initial velocity, z', is determined from the suvat phase. You can use a similar expression for velocity and acceleration as A is just some constant.

The vertical spring will not exhibit shm about its natural length (not given), rather it will exhibit shm about the equilibrium position -mg/k as you have
mx''' = -mg - kx
or
mx'' + (mg + kx) = 0
or where z = x + mg/k then
z'' + k/m z = 0
so shm with w^2 = k/m.

Thats the top level stuff. If youve questions about any details, youlll have to be specific and if youre referring to your working, it helps to see it.
(edited 3 weeks ago)
Oh I see so you can only use the let __ = Asin(wt+phi) argument when its in the form of y" = - k/m y as shm for a spring is w^2 = k/m.

Im just gonna focus on the video because tbh the 2 seemingly different methods are making me more confused.

Why does the video set up the equation mx'' = -mg - kx rather than mx'' = mg - kx (since the forces should act in opposite directions?) as otherwise surely in the equilibrium position: mx" = 0 = -mg - kx
so x = -mg/k rather than just x = mg/k so then following that z = -mg/k -x?

For that matter, I thought x was the total extension of the spring but apparently not? So does the following diagram I drew communicate what z = mg/k + x is?
Reply 3
Original post by mosaurlodon
Oh I see so you can only use the let __ = Asin(wt+phi) argument when its in the form of y" = - k/m y as shm for a spring is w^2 = k/m.
Im just gonna focus on the video because tbh the 2 seemingly different methods are making me more confused.
Why does the video set up the equation mx'' = -mg - kx rather than mx'' = mg - kx (since the forces should act in opposite directions?) as otherwise surely in the equilibrium position: mx" = 0 = -mg - kx
so x = -mg/k rather than just x = mg/k so then following that z = -mg/k -x?
For that matter, I thought x was the total extension of the spring but apparently not? So does the following diagram I drew communicate what z = mg/k + x is?

The solution of an ode like
x'' + w^2x = 0
is
x = Acos(wt) + Bsin(wt) = Asin(wt+phi) = ....
You must have covered this but if there is a right hand side so -mg in this case, this would correspond to a particular integral or you could transform the equation
x'' + w^2x = -mg
into
z'' + w^2z = 0
as above. Its shm about a different equilbrium.

On your diagram you dont assign a positive direction. More worryingly, you have two arrow heads, so which way is positive? In the video, its positive up, so gravity is (negative) down and the extension from the natural length, x, is downwards, so newton 2 says
ma = - weight + compression
and compression is -kx as x is negative and compression acts to increase x and counter weight.
My bad... Im actually entirely self studying this topic which is clearly not going so well..

I completely neglected the fact that x could be a vector (displacement) and not just the compression length (distance), and I was pretty confused when you initially said "so shm about the equilibrium position x= minus mg/k", but actually thinking about it now it almost appears obvious as you want the new transformed point to be 0 at the equilibrium point and since x = -mg/k at this point it shouldve jumped out.

Just to check my working:
z" + k/m z = 0
z = Asin(wt+phi), z = mg/k + x
when t = 0, x = 0, z = mg/k
Asin(phi) = mg/k
when t = t_2, x" = 0, x = -mg/k, z = 0
sin(wt_2+phi)=0
wt_2+phi = 0

The "someone elses working" I posted originally actually used the incorrect ma = mg - kx, which did not help fixing my confusion, but they managed to get the right answer anyways.
(edited 3 weeks ago)
Reply 5
Original post by mosaurlodon
My bad... Im actually entirely self studying this topic which is clearly not going so well..
I completely neglected the fact that x could be a vector (displacement) and not just the compression length (distance), and I was pretty confused when you initially said "so shm about the equilibrium position x= minus mg/k", but actually thinking about it now it almost appears obvious as you want the new transformed point to be 0 at the equilibrium point and since x = -mg/k at this point it shouldve jumped out.
Just to check my working:
z" + k/m z = 0
z = Asin(wt+phi), z = mg/k + x
when t = 0, x = 0, z = mg/k
Asin(phi) = mg/k
when t = t_2, x" = 0, x = -mg/k, z = 0
sin(wt_2+phi)=0
wt_2+phi = 0
The "someone elses working" I posted originally actually used the incorrect ma = mg - kx, which did not help fixing my confusion, but they managed to get the right answer anyways.

The extension or displacement x certainly has a sign and the basic spring equation
tension = -kx
only works if x is a positive displacement and tension is therefore negative and acts to decrease x and accelerates (the particle) back to the center. When x is negative and the spring is in compression
compression = -kx
so compression is positive and acts to accelerate the particle back to the center. This is the central feature of shm/springs. I noticed the typo in the posted workings, but youve got to understand the basic stuff and set up the equations correctly, not just be able to read through someone elses correct or incorrect working.

In your working, you also have the initial condition
x'(0) = z'(0) = whatever the velocity is from the suvat equation
so two initial conditions from which you can solve for A and phi. So that means you have the solution for z (and x). Then use that equation to solve for the necessary times. Ive not worked through this part, but that seems to be the basic approach.

Note the Acos(wt) + Bsin(wt) form is arguably easier to transform the initial conditions into A and B and then convert that into the Ccos(wt+phi) or ... I think there was an isaac question which asked about the basic forms/solutions of shm.
(edited 3 weeks ago)
Original post by mqb2766
The extension or displacement x certainly has a sign and the basic spring equation
tension = -kx
only works if x is a positive displacement and tension is therefore negative and acts to decrease x and accelerates (the particle) back to the center. When x is negative and the spring is in compression
compression = -kx
so compression is positive and acts to accelerate the particle back to the center. This is the central feature of shm/springs. I noticed the typo in the posted workings, but youve got to understand the basic stuff and set up the equations correctly, not just be able to read through someone elses correct or incorrect working.
In your working, you also have the initial condition
x'(0) = z'(0) = whatever the velocity is from the suvat equation
so two initial conditions from which you can solve for A and phi. So that means you have the solution for z (and x). Then use that equation to solve for the necessary times. Ive not worked through this part, but that seems to be the basic approach.
Note the Acos(wt) + Bsin(wt) form is arguably easier to transform the initial conditions into A and B and then convert that into the Ccos(wt+phi) or ... I think there was an isaac question which asked about the basic forms/solutions of shm.

Sorry to jump in but I was running an eye through this question and have a doubt. When the ball comes to equilibrium, velocity is 0, hence shouldn't you be able to do 0=omega*A*cos(omega*t+phi) therefore omega*t+phi=pi/2 (as that is when cosine is 0) and then solve for t? That is because we have taken x=A*sin(omega*t+phi), however shouldn't it not matter if I had taken x(t)=A*cos(omega*t+phi) and then said that v=0 at equilibrium point so 0=-omega*A*sin(omega*t+phi)?
(edited 3 weeks ago)
Reply 7
Original post by Javier García
Sorry to jump in but I was running an eye through this question and have a doubt. When the ball comes to equilibrium, velocity is 0, hence shouldn't you be able to do 0=omega*A*cos(omega*t+phi) therefore omega*t+phi=pi/2 (as that is when cosine is 0) and then solve for t? That is because we have taken x=A*sin(omega*t+phi), however shouldn't it not matter if I had taken x(t)=A*cos(omega*t+phi) and then said that v=0 at equilibrium point so 0=-omega*A*sin(omega*t+phi)?

Not worked it through fully (or read the previous solution or fully understand what youre saying), but the velocity would be zero when x is at the minimum (so the spring is maximally compressed). This would not be an equilibrium as the acceleration would be maximal. The overall time would be twice the time to get to the point where x is minimised.

Whether you do
x = Asin()
or
x = A cos()
is fairly irrelevant as it corresponds to different phi.
Original post by mqb2766
Not worked it through fully (or read the previous solution or fully understand what youre saying), but the velocity would be zero when x is at the minimum (so the spring is maximally compressed). This would not be an equilibrium as the acceleration would be maximal. The overall time would be twice the time to get to the point where x is minimised.
Whether you do
x = Asin()
or
x = A cos()
is fairly irrelevant as it corresponds to different phi.
True you will get different phi thanks!
Reply 9
Original post by javier garcía
True you will get different phi thanks!

Agreed, though Ive worked it through (roughly) now and there are a few strange things in the notes.

Assuming positive up (x and z), then the initial velocity is down or negative so it would "make sense" to do
z = Acos(wt + phi)
as then (A>0), this would correspond to a 0<phi<pi/2 as this produces an initial positve displacement and negative velocity. If |phi|>pi/2 you have to be careful when doing the arctan(). The arctan(phi) is relatively simple as the numbers (h, m) are chosen to be related. Then its just a case of 2pi-2phi and ... Its(obviously) analogous to converting sin+cos into harmonic form.

(Both of you) Id really suggest working through it clearly and largely ignore the notes. Theyre mostly right, though have a few "typos" and really the learning is about how you set up the equations.
(edited 3 weeks ago)
actual math 1.jpg
math 2.jpg

I managed to do it using Acos.. and Asin… but am not getting anywhere with Acoswt + Bsinwt

edit: nvm I made a really stupid mistake - it should be tan(wt) = wmg/kv, so yes you get the same result
(edited 3 weeks ago)
Original post by mosaurlodon
Attachment not found

Attachment not found

I managed to do it using Acos.. and Asin… but am not getting anywhere with Acoswt + Bsinwt

Looks about right and you noticed the negative phi for Asin() so how does that relate to time? Its also worth thinking about what circular motion and phi represents.

You could imagine it either as a sinusoidal wave or a "circle" (z and z' being the x and y axes) and the value of A isnt important for this calculation. For the cos one,
wt = phi
represents the time taken to reach the initial state if the initial conditions had been the more usual z(0)=A and z'(0)=0. It also represents the time taken to return to {z,z'}={A,0} after the ball as left the spring. So the time it is in contact with the spring is simply the time for a full cycle - these two sections so
(2pi - 2phi)/w
Id argue its easier to do that than splitting the half cycle into two portions, calculating and summing them and then doubling. Its also about ~2/3 page of working, with a good sketch, so 1) suvat drop 2) set up the z ode, 3) get w and phi 4) T = (2pi - 2phi)/w

Whether you do Acos+Bsin or Ccos or Dsin, its worth being clear about what the motion actually is in order to interpret the required time correctly. For the Dsin() one, youve got phi=-1 and this only makes sense if A is negative as well as you want z positive and z' negative. Then youd have to think about which portion of the cycle youre calculating the time for, and for this case, what a negative time means.
(edited 3 weeks ago)
For your middle paragraph, (im still thinking about the other points Ill get back to you i.e. "negative phi for Asin() so how does that relate to time? Its also worth thinking about what circular motion and phi represents." and "what a negative time means.")


more math 1.jpg



I visualised phi as the 'lag' for each half period as since T = 2pi/w with the total lag T = 2pi-2phi/w - which is a nice shortcut I had no clue about.
(edited 3 weeks ago)
Original post by mosaurlodon
For your middle paragraph, (im still thinking about the other points Ill get back to you i.e. "negative phi for Asin() so how does that relate to time? Its also worth thinking about what circular motion and phi represents." and "what a negative time means.")
more math 2.jpg
-> have I drawn incorrectly?
more math 1.jpg
I visualised phi as the 'lag' for each half period as since T = 2pi/w with the total lag T = 2pi-2phi/w - which is a nice shortcut I had no clue about.

I think youre right, though not 100%. Using Dsin implies the basic initial condition is z(0)=0 and z'(0)=wD. Here we have a negative initial velocity so D<0 (or |phi|>pi/2) and phi=-1. The sum of the two phi's (magnitudes, cos and sin) is pi/2 and in this case (sin) phi represets the wt interval from the time the ball drops on the spring until z=0, though unless you sketch it, its easy just to follow someone elses working.

your sketch for cos is about rigt but the dashed line should be down a bit and if youre plotting against t then
Ccos(3t+0.57) = Ccos(3(t+0.19))
so a left shift of 0.19 and a scaling of 3, so the interesting time is from 0 to ~1.6 and the period of the cos wave from -0.2 to 1.8.

Do something similar for the sin one, as these are the pictures you have to have in your head to understand the question
I double checked them with desmos and logic of my original workings so these should hopefully be right

final math.jpg

I was playing around with the graphs in desmos and I figured out why the |phis| sum to pi/2.
Ccos(w(t+phi1/w)) and Dsin(w(t+phi2/w)), phi1 = 0.6, phi2= -1
You move the cos graph |phi1|/w to the left and sin graph |phi2|/w to the right
so the phase difference is |phi1|+|phi2|/w
and to match a cos to sin or vice versa you need to have a phase difference of pi/2 and scale by -1 so -1*f(x)
the graphs are already scaled by -1 as the A in the sin = -A of the cosine and since both graphs are squished by w in the t axis, the required phase difference is only pi/2w.
so |phi1|+|phi2|/w = pi/2w
|phi1|+|phi2| = pi/2

Thanks a lot :smile:
(edited 3 weeks ago)
Original post by mosaurlodon
I double checked them with desmos and logic of my original workings so these should hopefully be right
final math.jpg
I was playing around with the graphs in desmos and I figured out why the |phis| sum to pi/2.
Ccos(w(t+phi1/w)) and Dsin(w(t+phi2/w)), phi1 = 0.6, phi2= -1
You move the cos graph |phi1|/w to the left and sin graph |phi2|/w to the right
so the phase difference is |phi1|+|phi2|/w
and to match a cos to sin or vice versa you need to have a phase difference of pi/2 and scale by -1 so -1*f(x)
the graphs are already scaled by -1 as the A in the sin = -A of the cosine and since both graphs are squished by w in the t axis, the required phase difference is only pi/2w.
so |phi1|+|phi2|/w = pi/2w
|phi1|+|phi2| = pi/2
Thanks a lot :smile:

Mathematically that sounds comprehensive. I just thought that the difference between the two initial points (C,0) and (0,D) is pi/2 and the actual iniital point (mg,-v) makes an angle phi1 with one axis and phi2 with the other so ... I started to do this
https://www.desmos.com/calculator/wsngxkcb7q
which was meant to represent the Ccos() view and the missing arc at the top is 2phi (note the C,w scalling is missing but ...) It could be annoted somewhat but it was to try and get the idea betwen circular motion (with x=z' and y=z) and shm and really the latter is just a stretched (ellipse) version of the former and thinking about it in state space z,z' can make the initial conditions, phis, ... clearer but the sinusoidal curves are very useful.
(edited 3 weeks ago)
Thank you for the graph! Ill try to keep that intuition in mind for future problems, I also think the sinusoidal relationship will be a very handy tool to check my logic.

If I could ask one last question:
For the video method, it calculates t3 using T= 2pi/w and then quarters this.
Is the justification for this because in the graph that I drew, the only region that is a full "U-shape" is either side of t3? Thus to get the time of half of the full "U-shape" you just do T/4?

So a fast-pass method would just be subtracting the lag times of phi, one for going down and one for going back up.
So the quick way would be T_quick = 2pi-2phi/w which allows you to skip calculating 2(t2+t3)?
Original post by mosaurlodon
Thank you for the graph! Ill try to keep that intuition in mind for future problems, I also think the sinusoidal relationship will be a very handy tool to check my logic.
If I could ask one last question:
For the video method, it calculates t3 using T= 2pi/w and then quarters this.
Is the justification for this because in the graph that I drew, the only region that is a full "U-shape" is either side of t3? Thus to get the time of half of the full "U-shape" you just do T/4?
So a fast-pass method would just be subtracting the lag times of phi, one for going down and one for going back up.
So the quick way would be T_quick = 2pi-2phi/w which allows you to skip calculating 2(t2+t3)?

Yes (I think). The Csin() phi represents the arc(s) above the x-axis, so the length of time until the velocity is maximum so wt+phi = pi/2. Then the 1/4 (2pi/w) is the quarter time period so the length of time from crossing the x-axis until reaching the minimum (zero velocity)) of Csin(). The key reason for this is wt is a linear function of time, so it takes T/4 to complete a quarter of the cycle.

As I said at the start, its easy to use the hints/worked solutions too much and easy to skip over this (important) understanding. Indeed in this question, its arguably more complex than necessary and it was a bit easier to subtract 2phi from T using the Dcos() representation.
(edited 3 weeks ago)

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