Question

If a simple pendulum has significant amplitude (upto a factor of 1/e of the original) only in the period between t=0 to t=ms then m->avg. life. Then what is the avg. life time of pendulum under a damping force (-bv) ?

Answer 1

@ganeshie8 @samigupta8 @parthkohli @mayankdevnani

Answer 2

for the DE for damped SHM for pendulum i get \(\ddot \theta +\frac{b}{ml} \dot \theta + \dfrac{g}{l} = 0\)

with roots for characteristic equation

\(-\frac{b}{2ml} \pm \sqrt { \left ( \frac{b}{2ml} \right) ^2 -\dfrac{g}{l} }\) .....but check that, BTW

\(= -\alpha \pm \beta\)

and for under damped/ oscillating shm \(\theta = e^{- \alpha t} ( A \sin \beta t + B \cos \beta t )\)

and then i'd just make up my own IV's to discount the cosine term, then play with the sine term in the brackets but focus on the amplitude decay, so play with \( A e^{- \alpha t} = e^{-1}A\)

well, that's the broad idea :-)

Answer 3

I directly applied the formula A=A0e^(-bt/2m).. and i got bt/2m = 1.. now for m sec..let it be 'n' so that it doesn't get confusing..! so for n sec.. should i do.. n=2m/b ? @IrishBoy123

Answer 4

\(\large A_o\color {red}{e^{-1}}=A_oe^{-bt/2m \color{red}{l}}\) , n'est-ce pas?

so that's \( -1 = -\dfrac{b t}{2ml} \Rightarrow t = \dfrac{2ml}{b} \)

for clarity, i made the opening DE to be : \(m l^2 \ddot \theta +bl \dot \theta + mgl \theta = 0\) and took it from there [[spotted a missing \(\theta\) in my opening post, first line :-(( ]]

i think the algebra follows correctly from that , but yo may have strated somewhere different, so just check.....but that seems pretty good attempt....

Answer 5

i am gettin 2/b but answer 1/b..

Answer 6

i'm sticking with the answer posted ( :-)) ) but please do share if there is something awry here

good luck!

Answer 7

you mean 2ml/b?..but l is not given in the Q..

Answer 8

\(l\) has to be in,

.....the original DE i used is \(m l^2 \ddot \theta +bl \dot \theta + mgl \theta = 0\)

Answer 9

actually, i think i agree with the point about "l" ..... :-((, mea culpa

taking moments for the pendulum about the pivot gives:

\(I \alpha = - l \times bv - l \times mg \sin \theta \)

\(\implies ml^2 \ddot \theta = - l \times b l \dot \theta - mgl \sin \theta \)

\(\implies \ddot \theta + \dfrac{b}{m} \dot \theta + \dfrac{g}{l} \sin \theta = 0 \)

\(\implies \ddot \theta + \dfrac{b}{m} \dot \theta + \dfrac{g}{l} \theta = 0 \)

so it's \(exp ( -\frac{b}{2m} t)\)

so \( t = \dfrac{2m}{b}\)

Answer 10

so my answer was right!

Answer 11

Thanks @IrishBoy123