Simple Harmonic Motion Derivations using Calculus (Mass-Spring System)

[@WarriorBane]

Hello sir, I just wanted to let you know I appreciate your commitment to these videos

[@qwertyuiop-ru4ii]

Have my physics exam in 2 hours thanks for the awesome video

[@FlippingPhysics]

Best of luck!

[@tannerwalston5868]

Great videos man just wanted to say I appreciated the effort you put in, thank you!

[@johnedison5409]

Sir I love u alot love u love love u....sir r the scientists.......make more vidoes ....

[@cupostuff9929]

How do you derive that condition for SHM

[@iremnurbulut3464]

Sir THANK YOUUU

[@naravishthongnok5899]

I still have no idea why my prof. never showed us a proof for this equation, All he did was math handwaving.

[@cupostuff9929]

This was pretty good

[@veco1174]

this is perfect.

[@colchi2547]

This was awesome !

[@puns-here]

someone help me through this; shouldn't omega in these equations be angular velocity instead of angular frequency? its a little confusing

[@bluemesastudios]

Double check my reasoning here, but I believe in this case, since we are talking about the motion of a spring, we talk about the "frequency" of its periodic motion, whereas when we speak of angular velocity, we are in the context of rotational motion - like a spinning top. I defer to others more experienced in the subject matter : )

[@kon7693]

Just curious does this derivation apply with a vertical spring? As now the net force equation needs to account for the force of gravity?

[@FlippingPhysics]

Yes, changing to vertical simply adjust where the equilibrium position is. www.flippingphysics.com/mass-spring-horizontal-vertical.html

[@Paul-wh7nb]

To train for conditions during a manned mission on the surface of Mars, where the acceleration due to gravity is only 3.72 m/s^2, astronauts can be hung on a spring harness while on Earth. To provide the necessary force, we can stretch a stiff (large k) spring by a little, or a softer (small k) spring by a lot. Should we use a stiff or soft spring to best simulate the conditions on the Mars surface as the astronaut hops around the training ground? Also can you help with the equations of motions for this question?

[@carultch]

I would expect that you would want to use a soft spring stretched a lot, rather than a stiff spring stretched a little. Take a look at the gravity rig that Mythbusters set up for a similar experiment: kzbin.info/www/bejne/jZnOe2OneZllfNE You notice that they have a high ceiling to support the rig with the bungee cords, and Adam has to climb up on a ladder to attach his harness when the cords are unloaded. Adam is wearing enough equipment to make his body weigh 360 pounds prior to attaching to the bungee cords, which is what Neil Armstrong in his spacesuit would weigh on Earth. He is using the bungee cords to support 300 pounds of the weight of Adam + equipment, so that the floor applies 60 pounds of normal force, which is what the Neil Armstrong weighed on the moon. The more he is able to stretch the cables as he climbs back down off the ladder, the greater the range of motion he can have within a reasonable tolerance of "weighing" 60 pounds.

[@Paul-wh7nb]

Thank you for the detailed reply Carl. That video made so much sense into understanding the whole concept. I have 2 more questions to ask 1. How does the spring force change as Adam or an astronaut in this case, hop up and down? 2. How do we ensure that the spring is “most similar” to moon/Mars?

[@carultch]

@@Paul-wh7nb Glad I could help. To answer your additional questions: 1. If the bungee cords were fixed in place on the top of the rig, the spring force would vary linearly with the position of Adam's harness, assuming the bungee is linear-elastic. The stiffer the bungees, the greater the value of dF/dz, where z is the position of Adam's harness. They didn't exactly show the details of the rigging at the top of the cords, so something else could be going on behind the scenes to selectively deploy different lengths of cord, depending on his height. 2. We'd like to minimize dF/dz as much as as possible, which means the k-value of the bungee cords need to be minimized, and make the range of body movements small, relative to the initial change in length of the bungee cord at the neutral standing position. To apply numbers, consider a 100 kg astronaut, and a desired 372 Newton normal force, in his neutral standing position, 1 meter above the floor. Suppose the range of motion is 0.5 m to 1.5 m, and the tolerance on the normal force is +/-10%. This means the k-value of our bungee is 74 N/m. To support the astronaut at the neutral 1m harness position, the cord would be stretched 8.2 meters from original length, which will is 8.7 meters of stretching at the kneeling position. A bungee cord can stretch to twice its original length, so this would mean an 18 meter bungee cord for its relaxed length. It's unrealistic to have a room with a 27 meter ceiling to support both the 18 meter initial length, and 8.7 meter change in length, which is skyscraper height. A 10 meter ceiling is more realistic. So this would mean that behind the scenes, there is a system of pulleys to absorb the excess length. A serpentine path through multiple fixed pulleys inside the ceiling unit.

[@carultch]

@@Paul-wh7nb You also hear Adam's comment about how he is painfully aware of the 300 pounds of tension in the cables, during the experiment. This is a force you wouldn't experience when on the moon or Mars, in the low gravity. No matter what you do, the force from the bungees will be locally applied to the astronaut's body, and will not eliminate the experience of being in Earth's gravitational field. This is why it is common to use underwater pools, rather than overhead bungee rigs to train astronauts. You can use the buoyancy force to support most of the astronaut's weight, which is much better distributed than the harness. Still not uniformly distributed on every kg, like gravity, but at least it is distributed throughout the surface of the astronaut's body, instead of concentrated at the harness.

[@mohfa1806]

Thank you for your great videos i have question please : What if we assumed that the acceleration (a) is in the opposite direction of (postive x) then when we apply newton's second law projecting on the positive (x direction) we would get the following equation : - kx = - ma ma - kx=0 Which is different differential equation than what you derived ? Whats wrong with my logic ? , i am confused ! Thank you in advance

[@robertmorrison1657]

We know that the force applied is -kx due to the spring. It will be equivalent to the mass multiplied by the acceleration, not the negative of the mass times the acceleration. Thus: -kx=ma. If acceleration is negative, then x is positive.

[@mohfa1806]

@@robertmorrison1657 thx for the reply... But when we apply newton's second law aren't we projecting the spring force and acceleratiin along a negative or positive x-axis ?? , at some time in oscillation both spring force and acceleration are in the negative direction and applying newton's second law projecting on the positive x axis will lead us to : - kx = - ma ? , this is where iam confused

[@robertmorrison1657]

@@mohfa1806 You are correct when you say that spring force and acceleration are projected along an x-axis with positive and negative values. Whenever the spring force is in the negative direction, the acceleration will always be negative. Whenever spring force is positive, the acceleration will also always be positive. With the equation -kx=ma, that will always hold. With the equation -kx=-ma, this does not hold, since whenever spring force is positive, acceleration will be negative. Also, when spring force is negative, acceleration will be positive.

[@mohfa1806]

@@robertmorrison1657 thank you for discussing this ussue with me , but when we project on the positive x axis and when we asuume that both the spring force and acceleration vectors are in the negative direction with | magnitudes | equal to (kx) and (a) repectively , then when we apply newton's second law along the positve x axis we will get : - kx = - ma , and it completely holds since in this case both (x) and (a) are always positive (as they are magnitudes) , so this will end us with the differential equation : ma - kx = 0 Which is different than the SHM differential equation !! , that is my original question and i am still confused why we can get two different valid differential equation for each assumption of the acceleration vector being positive or negative ?

[@seymourfroggs]

I'm sorry to say I don't like this. It's always too neat just to say omega squared just happens to be our constant when omega is defined for a separate idea. There is a better way - but I need your help - otherwise I wouldn't be here - and it is knowing from straight maths that the second derivative of sinx is -sinx and noticing that fits in here. This gets us to rotational motion without *assuming* it.

[@osayieseosa]

I think I'd stick with my textbook than this video

[@FlippingPhysics]

Sounds good. Enjoy your textbook!