Question

How do I solve this? Please help.

Answer 1

isnt this math?

Answer 2

no, it's physics :p


em and i worked this through on the maths forum so i am going to post a different [non-ideal ] solution here .... without fear of being accused of feeding an answer......




for string of mass m with linear density \(\lambda\), ie \(\lambda = \frac{m}{l}\).

we have \(L = T - V\\ T = \frac{1}{2} \; (l \lambda) v^2 \\ V = V_0 - (\lambda x) g \frac{x}{2} = V_0 - \lambda g \frac{x^2}{2}\)

\(\dfrac{\partial L}{\partial v } = l \lambda v\)

\(\dfrac{d}{dt} \left( \dfrac{\partial L} {\partial v } \right)= l \lambda \dot v = l \lambda \ddot x\)

\(\dfrac{\partial L}{\partial x } = \lambda gx\)

\(\implies l \lambda \ddot x = \lambda gx\)

\(\large \ddot x = \dfrac{g}{l} x\)

the auxiliary equation is \((D^2-\dfrac{g}{l}) = 0 \); so the solution to this is \(x(t) = A e^{\omega t } + B e ^{- \omega t}\) where \(\omega = \sqrt{\dfrac{g}{l}}\)

from the IV's: \(\dot x_0 = 0, x_0 = a\)

\(\dot x =\omega A e^{\omega t } - \omega B e ^{- \omega t}\)
\(\dot x_o = 0 \implies A = B\)

\(x_o = a \implies a = A + B \implies A = B = \dfrac{a}{2} \)

\(x(t) = \dfrac{a}{2} e^{\omega t } + \dfrac{a}{2} e ^{- \omega t}\)

we are looking for T such that \(x(T) = l\) so

\(l = \dfrac{a}{2} e^{\omega T } + \dfrac{a}{2} e ^{- \omega T}\)

\(a e^{2 \omega T } - 2l e ^{\omega T} + a = 0\)

from the quadratic formula:
\(e^{\omega T}= \dfrac{2l \pm \sqrt{4l^2 - 4 a^2}}{2a} \\ = \dfrac{l \pm \sqrt{l^2 - a^2}}{a}\)

\(\omega T= \ln \left( \dfrac{l \pm \sqrt{l^2 - a^2}}{a} \right) \)

\( T= \sqrt{\dfrac{l}{g}} \ln \left( \dfrac{l \pm \sqrt{l^2 - a^2}}{a} \right) \)

[but i don't know how they chose between the - and + options...]