"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'm currently doing A levels in the UK and I havent seen a video that explained it in a way that I could derive it myself. Thanks a lot for showing me that it would be easy enough to derive if I forgot 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
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.
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.
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.
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
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
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)
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.
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.
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
you make me understood two topics in a row,spring and circuit topics,thanks!!
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.
That was pretty straightforward and simply explained.. Thanks ❤
I'm currently doing A levels in the UK and I havent seen a video that explained it in a way that I could derive it myself. Thanks a lot for showing me that it would be easy enough to derive if I forgot it!
Thank you for the clear and precise information. Also for the comparison you did with resistors.
great video man, my test is in an hour and this save me
Very clear! Thanks so much
what a fantastic video. Your explanation is so clear and simple so thank you!
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.
Man you expalined it so clearly. Thank u!
At first, I was blind, but now I see. Thank you good sir.
Great video, out professor teach us from it.
He doesn’t do shit
The video is too old but still up to date! Gonna help me always!
Thank you so much for making this content! Helped me with my homework when I was getting pretty confused!
Thank you man appreciate this video
this is so helpful
unlike our p.o.s book which doesn't even mention this topic yet they are in the papers
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!
Thank you so much! Not all heroes wear capes ;)
Thank you! Very clear explanation
Very very much helpful.thank you
JEE Aspirants assemble
Yuss surr!!
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).
really helpful :)
After years being in University, I am back to review and re watch your videos for my upper division course. Thank you !
It was crystal clear explanation.. thanks sir !
Great explaination thank you!
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.
Thank you
Learnt a lot Thank you
Thank you so much! I'm doing dynamics hw and we're now on the chapter about Work and energy!
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:-)czcams.com/video/45z6wwRMUsc/video.html
Like,share and subscribe my channel😊
Great vid.
Awww that's so cool ! Thank u !
Thank you Sir..... this really helped!
thanks for your beatiful explication
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.
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? 😄
Im in university and still watch your videos !
Good work 🙏sir
Brilliant👍👍👍
Loved it sir...
thank youuu ,you help me so much
THANK YOU SO MUCH!!!!!!
Thank You!
wow great video prof
Thank you,thank you ,thank you
Thank you sooo much Sir.
Keep up the good work.
Thank you ❤
Just divide both sides of the equation by F. It turns all of the F's up top into 1's.
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.
Amazing, you have cleared all of my doubts... thank you so much!
Great video man! You really helped me in understanding the why in this!
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.
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:-)czcams.com/video/45z6wwRMUsc/video.html
Like,share and subscribe my channel😊😊
great vid, thanks
If springs are in series and parallel connection which parameter remain constant for all the springs???
DUDE you are good.
THANK YOU
Thank you sir!
More related to capacitors than resistors? Is it a fair guess that this applies to compression springs as well?
Thanks Sir 👍
Short and simple
Thank you very much :3
thanks a lot. Made my day 😅😌😊☺😉😉😊😄
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
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
Great 👍..tq
Nice one 👌👌
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.
thank you!
What happen to the no of coils in parallel series would they be equal to Spring Stiffness ?
So, you're saying that the spring's constant depends on the length of the spring? Where that came from?
thank you so much
For compression do you just use a negative force?
thanks a lot
Thanks!
Thanks sir
Nice work
wow nice one
How the e is same as both spring.
Very good
Thankyou
Springs and spring oscillators are interestingly the inverse of electrical circuits.
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)
THANK YOU!!
So are you there??
Thank you...
Sense has been made...
Are you there after 8 years?
why is the force in each spring the same?
for series springs
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.
if we cut it in 4 parts so K will be four times original K ? is it righy
thanks alot mann !! , appreciate the video big time :)
Good One!
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.