Weighing at Sea
https://isaacphysics.org/questions/weigh_sea?board=1ee488dc-7873-4ac4-b103-07f6015a7dd4&stage=a_level
Im a bit lost for this question - I tried watching the video and creating diagrams to represent the max/min amounts of extension but I don't really know how to progress from here.
I get "x" is the amplitude we're trying to calculate but I dont know how to remove y1 and y2 - are they just proportional to m1 and m2, so that proportionality cancels out? Not too sure and the ideas Ive tried havent been getting me anywhere.
Help much appreciated.
Im a bit lost for this question - I tried watching the video and creating diagrams to represent the max/min amounts of extension but I don't really know how to progress from here.
I get "x" is the amplitude we're trying to calculate but I dont know how to remove y1 and y2 - are they just proportional to m1 and m2, so that proportionality cancels out? Not too sure and the ideas Ive tried havent been getting me anywhere.
Help much appreciated.
Reply 1
Working:
Original post by mosaurlodon
Working:
Looks like youre 1/2 there. You know
•
w^2 = 4pi^2/T^2
•
max acceleration is k(y2-y1)/(2m)
Reply 3
so
m = k(y1-y2)*T^2/(8*pi^2*x)
and x ∝ m and y ∝ m
so x = alpha*m, y1 = alpha*m1 and y2 = alpha*m2
I've tried to sub that in but tbh im still struggling to find out a cancellation for k and alpha - I may be overcomplicating it but is my working out right so far?
m = k(y1-y2)*T^2/(8*pi^2*x)
and x ∝ m and y ∝ m
so x = alpha*m, y1 = alpha*m1 and y2 = alpha*m2
I've tried to sub that in but tbh im still struggling to find out a cancellation for k and alpha - I may be overcomplicating it but is my working out right so far?
Original post by mosaurlodon
so
m = k(y1-y2)*T^2/(8*pi^2*x)
and x ∝ m and y ∝ m
so x = alpha*m, y1 = alpha*m1 and y2 = alpha*m2
I've tried to sub that in but tbh im still struggling to find out a cancellation for k and alpha - I may be overcomplicating it but is my working out right so far?
m = k(y1-y2)*T^2/(8*pi^2*x)
and x ∝ m and y ∝ m
so x = alpha*m, y1 = alpha*m1 and y2 = alpha*m2
I've tried to sub that in but tbh im still struggling to find out a cancellation for k and alpha - I may be overcomplicating it but is my working out right so far?
Yes, youre overcomplicating it. From the simultaenous equations you must have
m = k(y1+y2)/g
and ...
The result is fairly intuitive if you think about a mass ratio multiplying g and ..
Reply 5
Oh I think I see what you mean
I got a different equation for m which I presume is what you meant could be a typo.
m = k(y1+y2)/2g, k(y2-y1)=2m * -4*pi^2/T^2 * x
so x =
(m_1-m_2)*T^2*g/((m_1+m_2)*4*pi^2)
but this equation was wrong but when I multiplied by negative one it was right??
I tried to check for the missing minus sign in my working but I couldn't find it
I got a different equation for m which I presume is what you meant could be a typo.
m = k(y1+y2)/2g, k(y2-y1)=2m * -4*pi^2/T^2 * x
so x =
(m_1-m_2)*T^2*g/((m_1+m_2)*4*pi^2)
but this equation was wrong but when I multiplied by negative one it was right??
I tried to check for the missing minus sign in my working but I couldn't find it
(edited 11 months ago)
Original post by mosaurlodon
Oh I think I see what you mean
I got a different equation for m which I presume is what you meant could be a typo.
m = k(y1+y2)/2g, k(y2-y1)=2m * -4*pi^2/T^2 * x
so x =
(m_1-m_2)*T^2*g/((m_1+m_2)*4*pi^2)
but this equation was wrong but when I multiplied by negative one it was right??
I tried to check for the missing minus sign in my working but I couldn't find it
I got a different equation for m which I presume is what you meant could be a typo.
m = k(y1+y2)/2g, k(y2-y1)=2m * -4*pi^2/T^2 * x
so x =
(m_1-m_2)*T^2*g/((m_1+m_2)*4*pi^2)
but this equation was wrong but when I multiplied by negative one it was right??
I tried to check for the missing minus sign in my working but I couldn't find it
m2>m1 and amplitude is usually positive. Yes, typo in previous post as it has to be the average.
(edited 11 months ago)
Reply 7
Oh I see so in the resulting expression you always multiply by -1 to make amplitude positive or did I just make a mistake in my working who knows
Thank you very much for your help as well.
Thank you very much for your help as well.
(edited 11 months ago)
Original post by mosaurlodon
Oh I see so in the resulting expression you always multiply by -1 to make amplitude positive or did I just make a mistake in my working who knows
Thank you very much for your help as well.
Thank you very much for your help as well.
One thing is to try and treat isaac as being less about an algebra slog and more about understanding how the problem is set up/how to interpret the result. So m2>m1 and y is positive down and the max displacement, acceleration pairs (magnitude) occur at the peaks of the shm. So how does that work through from your original x ~ y1-y2?
(edited 11 months ago)
Reply 9
Oh so It wouldnt matter If I took the displacement as being positive downwards/ positive upwards which would get me a +/-ve end result because the amplitude is defined as the maximum DISTANCE (so scalar) from the rest position to the peak, so I always take the modulus of this answer as this is the correct way to interpret amplitude?
Original post by mosaurlodon
Oh so It wouldnt matter If I took the displacement as being positive downwards/ positive upwards which would get me a +/-ve end result because the amplitude is defined as the maximum DISTANCE (so scalar) from the rest position to the peak, so I always take the modulus of this answer as this is the correct way to interpret amplitude?
Kind of, but it would be good to understand how to get the positive answer in the first place.
Reply 11
erm im kinda confused tbh
so I would get a positive answer in the first place If I took up direction as +ve - is that what you mean?
or are you saying that since x ~ y1-y2 (i.e. x is -ve) and since I took down as +ve it suggests x is going up?
edit:
my original expression for x = (T^2*g/(4*pi^2*(y1+y2))) * (y1-y2)
but im not really seeing anything - only that I get a +ve answer If I initially took up direction as +ve
so I would get a positive answer in the first place If I took up direction as +ve - is that what you mean?
or are you saying that since x ~ y1-y2 (i.e. x is -ve) and since I took down as +ve it suggests x is going up?
edit:
my original expression for x = (T^2*g/(4*pi^2*(y1+y2))) * (y1-y2)
but im not really seeing anything - only that I get a +ve answer If I initially took up direction as +ve
(edited 11 months ago)
Original post by mosaurlodon
erm im kinda confused tbh
so I would get a positive answer in the first place If I took up direction as +ve - is that what you mean?
or are you saying that since x ~ y1-y2 (i.e. x is -ve) and since I took down as +ve it suggests x is going up?
so I would get a positive answer in the first place If I took up direction as +ve - is that what you mean?
or are you saying that since x ~ y1-y2 (i.e. x is -ve) and since I took down as +ve it suggests x is going up?
The problem occurs because shm works with the peaks/troughs
•
max (positive) displacement, min (negative) acceleration
•
min (negative) displacement, max (positive) acceleration
Its one of those things where its easy to chug through the algebra, but its good to think about why errors occur, even if theyre minor (sign).
(edited 11 months ago)
Reply 13
Ah I see! So thats why flipping the directions work - so when you calculate the "max positive acceleration" its really the min (negative) acceleration which makes it then turn to the max (positive) displacement
Thank you!
Thank you!
(edited 11 months ago)
Original post by mosaurlodon
Ah I see! So thats why flipping the directions work - so when you calculate the "max positive acceleration" its really the min (negative) acceleration which makes it then turn to the max (positive) displacement
Thank you!
Thank you!
The max positive acceleration (which you calculate) occurs when the displacement is min (negative), and vice versa, which is probably what you meant in the bold? It must be that because of negative sign in the w^2 when you relate acceleration to displacement.
(edited 11 months ago)
Reply 15
I was just saying that if you take up direction to be +ve rather than down direction to be +ve then the final answer for "x" would change sign.
You actually start off with the same equations since the change of sign with displacement and acceleration cancels out
k(y2-y1)/(2m) = max (positive) acceleration
k(y1+y2)/2g = m
but since a= - w^2 *displacement
if you change the direction to be up as positive (which makes the displacement negative) that changes the a to be positive
So if displacement is negative you end up with the right answer without having to remove the minus sign.
You actually start off with the same equations since the change of sign with displacement and acceleration cancels out
k(y2-y1)/(2m) = max (positive) acceleration
k(y1+y2)/2g = m
but since a= - w^2 *displacement
if you change the direction to be up as positive (which makes the displacement negative) that changes the a to be positive
So if displacement is negative you end up with the right answer without having to remove the minus sign.
(edited 11 months ago)
Original post by mosaurlodon
I was just saying that if you take up direction to be +ve rather than down direction to be +ve then the final answer for "x" would change sign.
You actually start off with the same equations since the change of sign with displacement and acceleration cancels out
k(y2-y1)/(2m) = max (positive) acceleration
k(y1+y2)/2g = m
but since a= - w^2 *displacement
if you change the direction to be up as positive (which makes the displacement negative) that changes the a to be positive
So if displacement is negative you end up with the right answer without having to remove the minus sign.
You actually start off with the same equations since the change of sign with displacement and acceleration cancels out
k(y2-y1)/(2m) = max (positive) acceleration
k(y1+y2)/2g = m
but since a= - w^2 *displacement
if you change the direction to be up as positive (which makes the displacement negative) that changes the a to be positive
So if displacement is negative you end up with the right answer without having to remove the minus sign.
Its easy to get signs confused when youre setting up the problem and the way they set up the
g +/- acceleration max
where acceleration max > 0 is a "good" way to do it.
Setting up a problem with a negative acceleration min is likely to introduce sign errors, though its a valid approach and arguably what the question wants.
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