It would work approximately the same, although a rubber band doesn't obey Hooke's law as well as a metal spring does. There is negative curvature to the graph of force vs strain for a rubber band, so there is a high k-value at low strains, and a lower k-value at higher strains. Rubber bands also exhibit hysteresis, which is a source of loss in attempting to use them as an example of SHM.
Suppose you knew the spring constant and had to solve for ‘x’ Then that would be mg/k which would give you a positive value but the displacement is negative, isn’t it? ( according to your sign convention)
If you've already accounted for the opposite directions by identifying that +x is downward, and the spring force is upward, you don't need to include the negative sign. It is accounted for by establishing this fact through other means.
sorry for being 2 months late but 9.81 m/s^2 is the value of acceleration due to gravity. It is a universal constant, meaning that it is accepted as a fact, so it can be used even if the question doesn't explicitly state it as a given.
@@probably_drew I wouldn't call it a universal constant. More like a global constant. The Big G is a universal constant, while the little g is a constant specific to our planet. The little g can take a value as low as 9.76 N/kg (highest mountain in Ecuador), and as high as 9.83 N/kg (north pole). Usually Physics classes teach you to default to a value that is close to the global average, such as 9.8 N/kg or 9.81 N/kg, unless it is otherwise specified. IMO, it is unrealistic to have more than 2 significant figures in this number, unless you know the geographic location where the problem takes place. There is a standardized value of 9.80665 N/kg, that is the basis for defining the kilogram as a force unit, as it is used on a conventional scale. This value rounds to 9.81 N/kg.
What if the spring is compressed not stretched, will it be okay to have a negative spring constant? Please notice my comment i will taking an exam tomorrow
That still is a positive spring constant, it's just that the displacement of the mass is negative, and the force is positive, instead of vice-versa as it is for tensile springs. One reason we don't use the spring in compression, is due to buckling, a phenomenon that doesn't happen in tension. Sticking to the tensile mode of deformation means we can avoid this being an issue.
Idk what i would do without this teacher.
Lanca, I am glad to be here for you.
wow this video literally saved my entire physics grade for my final. thank you very much my good sir
AMIT??
@@minicenn yessir
update i got a near perfect grade on my final all thanks to this video hell yeah
Rock on!
👏the amt of effort u put in these videos is just impressive
Glad you appreciate it!
Wow your are the teacher I always wanted
And now you have me as a teacher!
Good point about the negative signs.
You are the best. Very clear explanation, thx
You are welcome.
This totally saved me. Thank you!!!
You are welcome!
this is actually really great, thank u a lot
Thanks for the video
Good God thank you so much for the explanation!!!! Youre amazing and a total blessing
You are so welcome
this was really cool. thank you.
U haven't covered topic on mechanical properties of solids/fluids.Please do so.
do you happen to have a video about solving with oscillations rather than displacement?
I determine the spring constant of an oscillating mass-spring system in this video. www.flippingphysics.com/shm-demonstration.html
Hey..I didn't understand why we did not use negative sign for the other equation
Hey, what happens if I hang an object with a rubber band tied with the spring?
It would work approximately the same, although a rubber band doesn't obey Hooke's law as well as a metal spring does. There is negative curvature to the graph of force vs strain for a rubber band, so there is a high k-value at low strains, and a lower k-value at higher strains. Rubber bands also exhibit hysteresis, which is a source of loss in attempting to use them as an example of SHM.
what if the final length is not given and just the mass added?
Does this method also work if we hang a mass to COMPRESS the spring?
Yes.
Thank you... I am from Bangladesh 😁
You are welcome. I am from the United States of America. You live your life roughly 10 hours ahead of mine.
Suppose you knew the spring constant and had to solve for ‘x’ Then that would be mg/k which would give you a positive value but the displacement is negative, isn’t it? ( according to your sign convention)
If you've already accounted for the opposite directions by identifying that +x is downward, and the spring force is upward, you don't need to include the negative sign. It is accounted for by establishing this fact through other means.
Thank you so much for the video....and love the tie dyes! :)
You are so welcome!
excellent
This is without calculating the spring's weight right?
Yes. This assumes an "ideal" spring, which has no mass.
Why the a(y) equal 0 ?
Where did 9.81 come from?
sorry for being 2 months late but 9.81 m/s^2 is the value of acceleration due to gravity. It is a universal constant, meaning that it is accepted as a fact, so it can be used even if the question doesn't explicitly state it as a given.
@@probably_drew I wouldn't call it a universal constant. More like a global constant. The Big G is a universal constant, while the little g is a constant specific to our planet. The little g can take a value as low as 9.76 N/kg (highest mountain in Ecuador), and as high as 9.83 N/kg (north pole). Usually Physics classes teach you to default to a value that is close to the global average, such as 9.8 N/kg or 9.81 N/kg, unless it is otherwise specified. IMO, it is unrealistic to have more than 2 significant figures in this number, unless you know the geographic location where the problem takes place.
There is a standardized value of 9.80665 N/kg, that is the basis for defining the kilogram as a force unit, as it is used on a conventional scale. This value rounds to 9.81 N/kg.
THANK YOU!
What if the spring is compressed not stretched, will it be okay to have a negative spring constant? Please notice my comment i will taking an exam tomorrow
That still is a positive spring constant, it's just that the displacement of the mass is negative, and the force is positive, instead of vice-versa as it is for tensile springs.
One reason we don't use the spring in compression, is due to buckling, a phenomenon that doesn't happen in tension. Sticking to the tensile mode of deformation means we can avoid this being an issue.
Anybody else here because Mr Matalavage didn't explain anything?
11454（惊撅）
what the hell is this video?
Asuuuuuuuu