Vibration | Part 2 | Mechanical Engineering

[@abhilashdandhare922]

Excellent teaching sir...U teach in a manner that an even average student can easily understand the concept and solve the problem....keep uploading the videos✌️✌️

[@kamalakarpatil5088]

Superb explanation bhai ! Vibration Engineering is a real analytical science which is a recent new science on which our universe is operating. I had vibration engineering as my favourite subject in 1978, that time it was just recently added in Engg syllabus.

[@ameygirdhari8703]

Superb lecture, sir your efforts is really helpful. It clears the concept of vibrations in easy manner

[@AkashRajput-qb8pg]

Amazing teaching skills... knowledge to sbke pass hota h but usko baatna hr kisiko Ni aata....thnx for this video

[@raj4624]

Deepraj Sir :- @ 46:56 Wn = underroot SL + mg(L/4) / I and I at Hinged by parellel Adi’s theorem 7mL^2/ 48 aaraha hai

[@bhupeshkpandey4186]

what a concept explanation,,supper teaching skill,,god bless you sir

[@vineetuniyal9615]

Thanku sir you made my concept clear 🙂

[@ajoyprasad963]

Your knowledge is excellent Thank you sir

[@syedrahamthulla3806]

Very good lecture by you sir

[@syedrahamthulla3806]

Thanks for making this video sir

[@aarrjunumrikar5345]

सर आपने बोला था की मुमेंट्स का समेशन आय×अल्फा जितना होगा! और यहा एक ही मुमेंट है। मगर दोनो की डायरेक्षन एक ही है। तो आपने जो इक्वेशन लिखा है उसमे न्यूटन का लॉ फॉलो नही हो रहा है! और अगर वैसे लिखा जाय तो बीच मे निगेटिव्ह साइन आना चाहीये... क्युकी आय× अल्फा = मुमेंट. सो (आय×अल्फा) - मुमेंट =0. मगर आपने प्लस लिखा है और डायग्राम मे भी एक ही डायरेक्षन आ रहा है। प्लिज थोडा क्लियर करोगे? बाकी व्हिडीओज से बहोत फायदा हुआ! उनके लिये शुक्रिया।

[@kunjansingh5967]

Square root of ((12mg+30sl)/7mL)

[@raj4624]

kunjan singh under root 12(S+ mg) / 7mL

[@raj4624]

@ 46:56 Wn = underroot SL + mg(L/4) / I and I at Hinged by parellel Adi’s theorem 7mL^2/ 48 aaraha hai

[@mongalmoykarmakar283]

@kunjan Singh right ans

[@conan5377]

Very good explaimation.

[@ArmanAli-fg5di]

Excellent explanation 🙂🙂

[@KJ-oq3xr]

Sir, in my humble opinion, the final formula of natural frequency you have given for compound pendulum seems to be wrong. Moment of Inertia about "O" should not be mk^2 as it is not Moment of inertia about CG. Using parallel axis theorem I(about O) = I(CG) + ma^2 or I(about O) = mk^2 + ma^2 And Therefore W natural= √[ga/(k^2+a^2)] Correct me if I'm wrong

[@sumitparida5328]

Yeah you're right

[@tejusramakrishna2965]

you are right

[@syedrahamthulla3806]

How it will come bro can u tell to me

[@varad_paraswar]

Exactly..

[@Mukeshkumar-uu8lp]

thanks for the video lecture

[@PKaushikYATHIRAJ]

mass moment of inertia of a ring about in plane axis is 0.5m*r*r,please correct it 7:28 also for a disc it is 0.25m*r*r

[@HimanshuSingh-ii5ki]

arre wo center of gravity se bata rhen hain . tum jyada gyan na do

[@syedrahamthulla3806]

Gud mrng deepaj sir

[@anantsinghal8010]

awesome , keep it up & upload more videos

[@rajdeepchatterjee400]

Got clarified

[@_Ashish_Gupta_]

Intro was ❤

[@vipinsingh6146]

Thankyou sir

[@raj4624]

under root 12(S+ mg) / 7mL

[@ashirvadmeena]

Thanku so much , it's really helpful ✨

[@prahaladsingh9045]

Mja aa gya sir padh kar thanku so much or haa Aap Kabir singh ka sahid kapoor lagte ho sir😍😍

[@asthatewari2830]

😭😭😭😭itni bezzati kisi se ni suni

[@AkashRajput-qb8pg]

Kya majak h bhai

[@SanaShaikh-xb5mt]

awesome teaching!

[@vijaykaravadra2271]

vahhh sir mass moment of inertia ke bhi question solve ho gaye sath mee

[@sumitparida5328]

Best video

[@DoUKnowMee]

500 th like by me🎊🎉

[@vaibhavkumar9894]

sir aapne jo sawaal diya tha isme mere w=square root of(12mgL+30L^(2)s/(7mL^(2))

[@deeprajchandrakar3188]

Correct hai...very good

[@sushantakumarmajhi6705]

Kese I nikla

[@shubhamdhiman2091]

In place of 7 there is 16

[@raj4624]

under root 12(S+ mg) / 7mL

[@raj4624]

@ 46:56 Wn = underroot SL + mg(L/4) / I and I at Hinged by parellel Adi’s theorem 7mL^2/ 48 aaraha hai

[@nikhilverma127]

Which music (EDM) is used in the start of the Video?

[@madanlaljat1881]

Sorry my mistake sir but ans is correct is Wn=under root of 30s+12mg/7m aa rha h

[@nehachauhan5800]

Thanks a lot sir jii ap nhi pdhate to vehicle dynamics ka kya hota hmara

[@nehachauhan5800]

Sir ek bat btaiye na mass ki bjh se moment initially to nhi ayega na kyuki initially mass act kr rha vhi axis pe ti C.G. act kr rha h to perpendicular distance kaise milegi

[@nileshmotwani7796]

Hello Neha, when you are twisting it with an angle, the mass will still act perpendicularly, so there will be a perpendicular distance, if you want solution, can give me your mail id. I'll send solved question

[@syedrahamthulla3806]

@@nileshmotwani7796 sir can u send to my mail id sir answer for that question

[@hello15239]

Sir ek question h ss ratan ka tom vibration chapter 18.1ka e wo is method se solve nhi kr pa rhe sir can u please help

[@prasannajade4966]

Square root of (30s+3mg)/7m

[@raj4624]

Prasanna Jade under root 12(S+ mg) / 7mL

[@raj4624]

@ 46:56 Wn = underroot SL + mg(L/4) / I and I at Hinged by parellel Adi’s theorem 7mL^2/ 48 aaraha hai

[@mehtaraj2873]

@@raj4624 unde root me 30s+mg/7m pura underroot me bro

[@mehtaraj2873]

@@raj4624 L vala term ans me nahi aayega

[@mehtaraj2873]

@@raj4624 (30s+3mg)/7m

[@mihirsrivastava1113]

Why spring force is not equal to -k×(delta)???

[@MrAAMNNITAllahabad]

brother, it is equal to -K*x I think, you are confused with the torque. I think in that, the delta had appeared.

[@rohitbhosle6521]

Delta (here we use x as initial displacement) is intial displacement if we consider as positive then spring force magnitude will be (delta*k) but act in opposite in direction so rather than representing it as (-k*delta ) we show in fbd that k*delta force act in opposite to initial displacement

[@madanlaljat1881]

Sir' Wn= under root of 30s/7m answer aa rha h

[@yme2003]

Tough lg rha h🙄😪

[@nehachauhan5800]

Sir please reply me

[@037anupkumarpandey5]

12g/7l answer AA rhaa hai sir

[@raj4624]

Anupkumar Pandey under root 12(S+ mg) / 7mL

[@princeviswas3642]

Thankyou sir

[@RImojr]

Thank you sir

[@skjafarhossain5059]

Thanks sir

[@hrithikjhariya]

Thanks sir