"We are just gonna assume" 4:08 This was such happy moment. I was literally so confused like everyone was saying that the extension of both springs is same and ........ Now i got it
I really need a great help in covering all General Organic Chemistry Principles concepts! :/ would you please help! :-} Please suggest some thing, how do I remember the Name Reactions?
@@anjana5887 Sure! What worked for me was making reaction maps and filling them in over and over!! For example, you can make a reaction map for say addition reactions ( or google one!!) by starting with an organic reactant, then draw an arrow to the product and fill in the blank reagents. You should also write the name the reaction while you do this, try to visualize or say out loud the mechanism/arrow-pushing, and other important info like "syn" or "anti" etc. You can also list reactants, then fill in the blanks with predicted products or vice versa. The key is to do this a bunch of times and to really force yourself not to look things up in your book/notes too much! The repetition + having to recall the info from memory is what helped me most. Another thing I did to remember reaction names or reagents was to make silly rhymes or numonics. Like for OSO4 I remembered "oh so syn-ister" to remember it was syn addition of OH lol I hope this helps!! You got this :D
Although I don't think his explanation was the best at this part, he is correct. Consider the first equation he derived for Keff for springs in parallel. If you have two identical springs, you can rearrange this equation to be Keff = k/2, meaning that putting two identical springs in parallel, halves your spring constant. Inversely, cutting a spring into two identical springs will double your spring constant.
You are right. But if you stretch it farther to get it back to the same length as when it was longer and stretched, then the new "x" will not be equal, neither the half of the original "x" (when it was longer and stretched).
I really don't know if what you are saying is true, but are you agree with me that the "x" is not the length of the spring? x is the displacement from the equilibrium position.
I'm doing this problem where I am calculating the spring constant of a vertical rod with a mass at the center. But I'm thinking the springs are in parallel instead of series. The total deflection at the center will be the same on both sides of the mass. In the problem both ends are fixes so both deflections have to be the same. I guess we only count them in series if the deflections of the springs are different.
Muhammad Ahmed I think you have to multiply by √2. If we take the spring with stiffness k and split it in two parts, we get two springs each with stiffness k. Thus effective stiffness is half of k. Substitute into the formula for period we get 2π√(m/0.5k) which is 2π√(2m/k) which is just multiplying by √2
Sir, amazing video! But I have a question, if the difference between K's on springs in series is quite big (I mean if we have a very soft spring and a very hard spring) when you pull with a relatively small force one will be deformed and in that case how is -k1*x1=-k2*x2 ? Maybe my intuition is wrong..
Last part is little confusing. What is meant by cutting spring in half ? -> do you mean cutting spring's constant in half or what do you cut in half? Equation F=-kx does not have length anywhere in formula. How did you come up with k'=2k? I don't think that the "length" of spring matter anywhere when modelling springs, unless we talk about rods in tension which act like a spring. There, you may be right as k[rod]=AE/L and if you cut rod in half you increase stiffness twice, as you do by increasing area or young's modulus twice, but in this example this is very misleading and confusing.
In case if the force isn't applied in the middle of the bar (in case of parallel springs),say its applied at a point P which divides the line into a ratio a:b then how will we obtain the equation for K effective.? thnx btw i know the equation but i am unable to understand how they derived it..Help would really be appreciated thnx... (y)
When you only have half of the spring, for the same force, it’s only going to stretch half as much, because there’s only half as much of it to stretch.
If there are to springs of spring constant K1 and K2 and if same force is applied to both they have time period T1 and T2 respectively. What would be the time period if they both are connected in parallel. any relationship between Teffec and T1 and T2.
ramanjot singh ramanjot, The T1 and T2 you are calculating are from T= 2*pi*sqrt(m/k). If you substitute keff in for k, then you know that Teff = 2*pi*sqrt (m /(K1 +K2)) for parallel springs.
This physics is the same when the springs are compression springs being compressed by a weight. Though, as in the video, if the springs have different constants they will compress different amounts. The spring with the greater spring constant will compress less.
So imagine a mass is resting on a spring and it compressed it x amount. Then imagine another, identical spring below both the mass and the other spring, ie in series. Would the second spring compress the same as the one above it? I would guess it would because the mass in the second spring is similar to that on the first. Is that right? Intuitively this seems wrong as the mass is compressing 2 springs the same amount, and this seems like twice the force, which can’t be true?
@@Coneman3 If the springs are identical and of negligible mass, then each spring would compress the same amount. So if each spring has a k of 100 N/m the mass has a weight of 10N, then each spring would compress 0.1 m. But if they have different k's, say 100 N/m and 200N/m, then the first would compress 0.1 m and the second would compress 0.05 m, regardless of their order but assuming they are in series. See this video for a fuller explanation of why the force on the bottom spring is the same as the force on the top spring: ruclips.net/video/VXu2gatnMWE/видео.html&ab_channel=lasseviren1
Many thanks for the quick and detailed reply. My initial thought have been confirmed by you. A spring below a mass at steady state is simply a mass to anything below it, as if the spring wasn’t there. It just seems counterintuitive because spring compression seems to always follow the rule of being in proportion to the force acting on it, in this case the mass. But if springs in series ‘shared’ the load, it would in effect be like the weight of the mass was shared, when that would be impossible. A simpler way of thinking about it could be that if I was to hold a mass off the floor, then someone held me, they would be taking me plus all the weight. This all springs in series under the same mass experience similar loads. Similar because mass of springs above has to be added too. This is going to help in a product I am developing, so it’s not just a theoretical enquiry 😉
@@dhts_bk For springs in series the overall spring constant (effective spring constant) will be less than either of the other spring constants. That's because the stretch for the combination must be greater than the stretch for either of them.
how do you do the formula for springs in opposing series of 2 different sizes and rates. etc I have a .50kg/mm x 450mm long spring in a opposing series set up with a .60kg/mm x 70mm. lets say that the 450mm spring compresses to 400mm and that would put the 70mm spring at 20mm n length. as the longer spring compresses the shorter spring extends till springs seperate completely. What would the rate be of the springs before they separate? I used the k1 x k2 / k1+ k2 = rate I know this formula works stacking them on top of each other but when doing the opposing series, do I use the inverse and subtract from the longer spring rate?? thanks Daivha
What if the springs are on either side of the mass ? does that mean they are in series as well ?so the first spring moves x in compression and the other one moves the same x in tension.
As long as the plate, or whatever it is, connecting the two parallel springs does not rotate, both springs have to deflect the same amount. That is, the plate, the end of spring 1, and the end of spring 2 all deflect the same amount. In general, since k_1 is not equal to k_2, different forces will be developed in each spring.
can someone help me, I am doing mathematical models in engineering and I want to know if a rotary damper would have both spring scenarios (both series and parallel) but opposing.
I'm 13 and haven't been taught this in class, and was sent away to find it out for homework.. I GET IT KINDA THANKS TO YOU :D but I'm really confused, what is the eff?
I wouldn't worry if I were you, were doing this exact stuff in A-level (and it confuses me). But the eff is the K-eff i.e. effective the spring constant if the 2 springs in series (or parallel) are thought to be just a single spring.
I love the fact that you started by diving straight into the topic!...
i love the fact that you have Gon picture in your pfp!...
@@ganeshgalaxygg2549 no you are shit.
you make me understood two topics in a row,spring and circuit topics,thanks!!
Great video! I learned a lot!! So ready for my test tomorrow : )
howd the test go buddy?
Hope you did well!
How was the test? LMAO
Still in school?
U failed I know
The video cleared all my concepts. The derivation of the equations in the beginning really helped me alot. Thanks and great video
"We are just gonna assume" 4:08
This was such happy moment.
I was literally so confused like everyone was saying that the extension of both springs is same and ........
Now i got it
Thanks for being clear and straight to the point. I understood everything! Actually good content
Brilliant teacher with brilliant concepts.Best explanation!! Thanks buddy.Still watching in 2020.
Thank you for the clear and precise information. Also for the comparison you did with resistors.
saved my life for my lab report!! thank you :)
I really need a great help in covering all General Organic Chemistry Principles concepts! :/
would you please help! :-}
Please suggest some thing, how do I remember the Name Reactions?
@@anjana5887 Sure! What worked for me was making reaction maps and filling them in over and over!! For example, you can make a reaction map for say addition reactions ( or google one!!) by starting with an organic reactant, then draw an arrow to the product and fill in the blank reagents. You should also write the name the reaction while you do this, try to visualize or say out loud the mechanism/arrow-pushing, and other important info like "syn" or "anti" etc. You can also list reactants, then fill in the blanks with predicted products or vice versa. The key is to do this a bunch of times and to really force yourself not to look things up in your book/notes too much! The repetition + having to recall the info from memory is what helped me most. Another thing I did to remember reaction names or reagents was to make silly rhymes or numonics. Like for OSO4 I remembered "oh so syn-ister" to remember it was syn addition of OH lol I hope this helps!! You got this :D
Thank you so much! I'll use these spring rules during my FE exam this coming week.
The video is too old but still up to date! Gonna help me always!
what a fantastic video. Your explanation is so clear and simple so thank you!
At first, I was blind, but now I see. Thank you good sir.
great video man, my test is in an hour and this save me
That was pretty straightforward and simply explained.. Thanks ❤
this is so helpful
unlike our p.o.s book which doesn't even mention this topic yet they are in the papers
Great video, out professor teach us from it.
He doesn’t do shit
Thank you so much for making this content! Helped me with my homework when I was getting pretty confused!
Man you expalined it so clearly. Thank u!
Very clear! Thanks so much
Although I don't think his explanation was the best at this part, he is correct. Consider the first equation he derived for Keff for springs in parallel. If you have two identical springs, you can rearrange this equation to be Keff = k/2, meaning that putting two identical springs in parallel, halves your spring constant. Inversely, cutting a spring into two identical springs will double your spring constant.
TheSwedishMoose he is cutting the spring constant in half not the actual spring
Good explanation , for easy watch here:-)ruclips.net/video/45z6wwRMUsc/видео.html
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Thank you! Very clear explanation
JEE Aspirants assemble
Yuss surr!!
It was crystal clear explanation.. thanks sir !
Thank you man appreciate this video
You are right. But if you stretch it farther to get it back to the same length as when it was longer and stretched, then the new "x" will not be equal, neither the half of the original "x" (when it was longer and stretched).
Thank you so much! I'm doing dynamics hw and we're now on the chapter about Work and energy!
Very very much helpful.thank you
Thank you so much! I've never had a proper physics mechanics class before and I was clueless but it makes sense now. You are an amazing teacher!
wow great video prof
After years being in University, I am back to review and re watch your videos for my upper division course. Thank you !
Thank you so much! Not all heroes wear capes ;)
Im in university and still watch your videos !
Learnt a lot Thank you
Loved it sir...
Keep up the good work.
Just divide both sides of the equation by F. It turns all of the F's up top into 1's.
How would you find the period if there is a mass in between two springs? What would the period be if there were two parallel springs acting on a mass?
DMeloMan u slove the reaction by constraints equation bro
Good explanation , for easy watch here:-)ruclips.net/video/45z6wwRMUsc/видео.html
Like,share and subscribe my channel😊
Great explaination thank you!
Great vid.
I really don't know if what you are saying is true, but are you agree with me that the "x" is not the length of the spring? x is the displacement from the equilibrium position.
Good work 🙏sir
1:24 sure! But why is the force acting on each springs the same in the first place? You can't just assume that, it must be explained why.
Thank you sooo much Sir.
really helpful :)
I'm doing this problem where I am calculating the spring constant of a vertical rod with a mass at the center. But I'm thinking the springs are in parallel instead of series. The total deflection at the center will be the same on both sides of the mass. In the problem both ends are fixes so both deflections have to be the same. I guess we only count them in series if the deflections of the springs are different.
Amazing, you have cleared all of my doubts... thank you so much!
Awww that's so cool ! Thank u !
Thank you Sir..... this really helped!
Brilliant👍👍👍
Nice work
Great video man! You really helped me in understanding the why in this!
thanks for your beatiful explication
DUDE you are good.
If springs are in series and parallel connection which parameter remain constant for all the springs???
Thank you ❤
Short and simple
More related to capacitors than resistors? Is it a fair guess that this applies to compression springs as well?
Great video, I used it in my Physics class. Question: Is your name really Lasse Viren or are you just a fan of the famous runner? 😄
if the spring is cut into two equal halfs,how is the time period affected
Muhammad Ahmed I think you have to multiply by √2. If we take the spring with stiffness k and split it in two parts, we get two springs each with stiffness k. Thus effective stiffness is half of k. Substitute into the formula for period we get 2π√(m/0.5k) which is 2π√(2m/k) which is just multiplying by √2
Thanks Sir 👍
thank you.Can you send me if the period of motion of amass connected to two springs connected to each other in series and parallel
That's T=2(pi)sqrt(m/k) where k is the effective k of both springs that is derived in this video.
Thank you,thank you ,thank you
Can i use the method if one spring is torsional spring and the other is translational spring and i need to find the k eq of the tow springs.
Nice one 👌👌
Great 👍..tq
Sir, amazing video! But I have a question, if the difference between K's on springs in series is quite big (I mean if we have a very soft spring and a very hard spring) when you pull with a relatively small force one will be deformed and in that case how is -k1*x1=-k2*x2 ? Maybe my intuition is wrong..
The very soft spring will have a small k and large x and the hard spring will have a large k and a small x.
Thank you!!
thank youuu ,you help me so much
For compression do you just use a negative force?
Last part is little confusing. What is meant by cutting spring in half ? -> do you mean cutting spring's constant in half or what do you cut in half? Equation F=-kx does not have length anywhere in formula. How did you come up with k'=2k? I don't think that the "length" of spring matter anywhere when modelling springs, unless we talk about rods in tension which act like a spring. There, you may be right as k[rod]=AE/L and if you cut rod in half you increase stiffness twice, as you do by increasing area or young's modulus twice, but in this example this is very misleading and confusing.
great vid, thanks
Thank you sir!
THANK YOU
What happen to the no of coils in parallel series would they be equal to Spring Stiffness ?
In case if the force isn't applied in the middle of the bar (in case of parallel springs),say its applied at a point P which divides the line into a ratio a:b then how will we obtain the equation for K effective.? thnx btw i know the equation but i am unable to understand how they derived it..Help would really be appreciated thnx... (y)
Good sir.. But one doubt.. how the extension of the spring became half of the original extension when it cut into two equal half..
When you only have half of the spring, for the same force, it’s only going to stretch half as much, because there’s only half as much of it to stretch.
If there are to springs of spring constant K1 and K2 and if same force is applied to both they have time period T1 and T2 respectively. What would be the time period if they both are connected in parallel.
any relationship between Teffec and T1 and T2.
ramanjot singh
ramanjot, The T1 and T2 you are calculating are from T= 2*pi*sqrt(m/k). If you substitute keff in for k, then you know that Teff = 2*pi*sqrt (m /(K1 +K2)) for parallel springs.
Springs and spring oscillators are interestingly the inverse of electrical circuits.
Does the equal extension of each spring in series also apply to springs compressed by a weight?
This physics is the same when the springs are compression springs being compressed by a weight. Though, as in the video, if the springs have different constants they will compress different amounts. The spring with the greater spring constant will compress less.
So imagine a mass is resting on a spring and it compressed it x amount. Then imagine another, identical spring below both the mass and the other spring, ie in series. Would the second spring compress the same as the one above it? I would guess it would because the mass in the second spring is similar to that on the first. Is that right?
Intuitively this seems wrong as the mass is compressing 2 springs the same amount, and this seems like twice the force, which can’t be true?
@@Coneman3 If the springs are identical and of negligible mass, then each spring would compress the same amount. So if each spring has a k of 100 N/m the mass has a weight of 10N, then each spring would compress 0.1 m. But if they have different k's, say 100 N/m and 200N/m, then the first would compress 0.1 m and the second would compress 0.05 m, regardless of their order but assuming they are in series. See this video for a fuller explanation of why the force on the bottom spring is the same as the force on the top spring: ruclips.net/video/VXu2gatnMWE/видео.html&ab_channel=lasseviren1
Many thanks for the quick and detailed reply. My initial thought have been confirmed by you. A spring below a mass at steady state is simply a mass to anything below it, as if the spring wasn’t there.
It just seems counterintuitive because spring compression seems to always follow the rule of being in proportion to the force acting on it, in this case the mass. But if springs in series ‘shared’ the load, it would in effect be like the weight of the mass was shared, when that would be impossible. A simpler way of thinking about it could be that if I was to hold a mass off the floor, then someone held me, they would be taking me plus all the weight. This all springs in series under the same mass experience similar loads. Similar because mass of springs above has to be added too.
This is going to help in a product I am developing, so it’s not just a theoretical enquiry 😉
@@Coneman3 Love the analogy of a person holding a person holding something else. All the best with your product!
So, you're saying that the spring's constant depends on the length of the spring? Where that came from?
THANK YOU SO MUCH!!!!!!
So that’s what happened in a series, will it be same in a parallel? I mean will the overall spring constant be less stiff?
Oh no I just get it so it will be more
@@dhts_bk For springs in series the overall spring constant (effective spring constant) will be less than either of the other spring constants. That's because the stretch for the combination must be greater than the stretch for either of them.
lasseviren1 ok thank a lot！I believe I‘ll pass the exam！
Good explanation , for easy watch here:-)ruclips.net/video/45z6wwRMUsc/видео.html
Like,share and subscribe my channel😊😊
thanks a lot. Made my day 😅😌😊☺😉😉😊😄
how do you do the formula for springs in opposing series of 2 different sizes and rates.
etc
I have a .50kg/mm x 450mm long spring in a opposing series set up with a .60kg/mm x 70mm.
lets say that the 450mm spring compresses to 400mm and that would put the 70mm spring at 20mm n length.
as the longer spring compresses the shorter spring extends till springs seperate completely.
What would the rate be of the springs before they separate?
I used the k1 x k2 / k1+ k2 = rate
I know this formula works stacking them on top of each other
but when doing the opposing series, do I use the inverse and subtract from the longer spring rate??
thanks Daivha
Very good
why is the force in each spring the same?
for series springs
Thank you
wow nice one
Thank you!
if we cut it in 4 parts so K will be four times original K ? is it righy
What if the springs are on either side of the mass ? does that mean they are in series as well ?so the first spring moves x in compression and the other one moves the same x in tension.
Are you alive?
Thanks sir
sir practically it is not possible to have x1=x2=x
in case of parallel spring arrangement as if they have
different stiffness
How will you justify???
As long as the plate, or whatever it is, connecting the two parallel springs does not rotate, both springs have to deflect the same amount. That is, the plate, the end of spring 1, and the end of spring 2 all deflect the same amount. In general, since k_1 is not equal to k_2, different forces will be developed in each spring.
can someone help me, I am doing mathematical models in engineering and I want to know if a rotary damper would have both spring scenarios (both series and parallel) but opposing.
I'm 13 and haven't been taught this in class, and was sent away to find it out for homework.. I GET IT KINDA THANKS TO YOU :D but I'm really confused, what is the eff?
I wouldn't worry if I were you, were doing this exact stuff in A-level (and it confuses me). But the eff is the K-eff i.e. effective the spring constant if the 2 springs in series (or parallel) are thought to be just a single spring.
Shian Harris that is why i am here
Good explanation , for easy watch here:-)ruclips.net/video/45z6wwRMUsc/видео.html
Like,share and subscribe my channel😊😊
No, actually it's right. Try cutting a spring in half and you'll see that it requires more force to stretch it by the same amount.
Thank you very much :3
Thankyou
thanks a lot
Sense has been made...
Are you there after 8 years?