Problem 168 Archimedes' Principle

SHO when the rod bobs up and down

[isAiming]

this sir is still faithful to his business of physics, thanks for these fun lessons, which, thanks to their unusual format, are remembered almost forever!

[skulltech]

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[IBORTAD]

With all good health to you and yours.

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[zy-qlgl]

this legent is still around here after the video that blew up, amazing

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[josephyossi]

im 13 and i dont really understand 100% of the content but i think that this is world class. hat off

[Incognito-cbtw]

Work can be calculated for two cases.... ∆x<L-Lsub .... And ∆x>L-Lsub ... Partially and fully submerged cases respectively...

[dasadasanudas]

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[ummunafishasto]

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[Hey_Ananyasingh]

Hello professor Walter lewin i am ananya Kumari I read in class 9 those who try never give up this question is i am solve it this question is jee the cylinders and wooden area so i am solve your after lecture
Thank you so much 🥰🌈😊😊

Sir you deserve all good and wealth in this world because you love other is your self and you always help all student and support thank you so much 😊😊😊😊

[avery-uw]

dit verwart me, maar ik ben blij voor je

[DaryanThelongear-rdtm]

MR JEGR WAS MY SCIENCE TEACHER AND NOW HE IS MY BROTHERS SCIENCE TEACHER I HAVE BEEN SUBSCRIBED SINCE 2016

[ulfhaller]

Length of the submerged portion of the rod, L = 3 m.
Cross sectional area of the rod = A m².
Mass of the rod,  M, will be equal to the mass of the water displaced by it.
=> M = ρV = ρAL
Fᵦ = Buoyant force.
Fᵧ = Gravitational force.
At equilibrium: Fᵦ = - Fᵧ
Fᵧ = Mg = ρVg, Fᵦ = ρgAL.
A small displacement, x => Fᵧ = ρALẍ, Fᵦ = -ρgAx
Fᵧ + Fᵦ = 0
ρALẍ + ρgAx = 0
ẍ + (g/L)x = 0 (SHO).
ω₀ = √(g/L)
T = 2π√(L/g)
g = 9.81 m/s² => T = 2π√(3/9.81) ≈ 3.47 s

For question b, I will wait for your solution.
I can not see why the period should change.

[michaelkouzmin]

Dear Professor, I did not publish my solution for this problem as I considered several terms unclear:
1) How the rod is held in the upright position - a rope/chain connecting to the bottom? Does it provide any reaction when the rod is achieving its upmost position?
2) Were we supposed to consider this oscilation system as a damped one (water against timber, friction, viscosity etc.)?

[Soham-Bhadra]

Solution to Problem 168 :
(a) Let Mass of the Rod = M, Cross-Sectional Area = A, Length of Submerged Portion of Rod = L, Atmospheric Pressure = p{0}, Displacement from Equilibrium Position = x .
Now observe that by Force Equation we have
Mx" = p{0} . A + M g - (p{0}+10^{3} . g (L + x))A including pressure at bottom of rod . Note that 10^{3} Kg/m^{3} is density of water and g is the force of gravity .
Again by Equilibrium Depth Equation we have
M. g = 10^{3} . g LA .
Combining both of them we get
10^{3} . (A L x" + g A x) = 0
=> - (g/L). x = x"
=>T = 2. \pi . (L/g)^{1/2} which is approximately 3.4746 s .
Therefore the period of oscillation is 3.4746 seconds .

(b) Even if the Atmospheric Pressure changes by 5%, the period of oscillation will not change since the equation of T shows that its value is independent of the value of Atmospheric Pressure .

[Madhukar-vedf]

Sir i am a big fan of you, I am studying class in 9th, i can't wait to meet you in real. I hope you still alive where I was doing jee preparation ☺️ 👍

[Incognito-cbtw]

Atmospheric pressure has no effect on the time period....

[akbarsyedqasim]

For the love of physics❤ I fell for the subject just because of you Sir. Love from India