How to show that the rate at which the period changes with respect to temperature is kT/2?

Question

I have this exercise to solve:

Temperature and the period of a pendulum For oscillations of small amplitude (short swings), we may safely model the relationship between the period T and the length L of a simple pendulum with the equation $$ T = 2\pi\sqrt{\frac{L}{g}};\ \ \ (1) $$ where g is the constant acceleration of gravity at the pendulum’s location. If we measure g in centimeters per second squared, we measure L in centimeters and T in seconds. If the pendulum is made of metal, its length will vary with temperature, either increasing or decreasing at a rate that is roughly proportional to L. In symbols, with u being temperature and k the proportionality constant, $$ \frac{dL}{du} = kL;\ \ \ (2) $$ Assuming this to be the case, show that the rate at which the period changes with respect to temperature is $$ k\frac{T}{2};\ \ \ (3) $$

Could someone, please, help with this? Any clue will be greatly appreciated.

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This is what I've come to so far

Namely, I need to show that the derivative of T with respect to u is $T'(u) = k\frac{T}{2}$ or that the relation of the rate of change of the period to the period itself is $\frac{T'(u)}{T} = \frac{k}{2}$

First, I tried to find T'(u), so I had to express T as T(u) and the only possible way seemed to be via relation (2): $L'(u) = kL$, that is $L = \frac{L'(u)}{k}$, then:

$$ T(u) = 2\pi\sqrt{\frac{L}{g}} = 2\pi\sqrt{\frac{L'(u)}{kg}} $$

Then, by the chain rule:

$$ T'(u) = \frac{\pi * L''(u)}{\sqrt{\frac{L'(u)}{kg}}} = \sqrt{\frac{(\pi * L''(u))^2 * kg}{L'(u)}} $$

But this seems to lead nowhere, since when I divided T'(u) by T(u) I got:

$$ \frac{T'(u)}{T(u)} = \sqrt{\frac{(\pi * L''(u))^2 * kg}{L'(u)}} : \sqrt{\frac{4\pi^2 * L'(u)}{kg}} = \\ \sqrt{\frac{\pi^2 * L''(u)^2 * (kg)^2}{L'(u)^2*4\pi^2}} = \frac{kgL''(u)}{2L'(u)} $$

Which is definitely nowhere near the required result, i.e. relation (3)

Answer 1

You complicated too much. Just use chain rule: $$\begin{align}\frac {dT}{du}&=2\pi\frac1{\sqrt g}\frac d{du}\sqrt L\\&=2\pi\frac1{\sqrt{g}}\frac 1{2\sqrt L}\frac{dL}{du}\\&=2\pi\frac1{\sqrt{gL}}\frac 12 kL\\&=\frac{kT}2\end{align}$$

Answer 2

Take logs of both sides: $$\ln T = \ln\left(2\pi\sqrt{\frac{L}{g}}\right) = \ln 2\pi - \frac{1}{2}\ln g + \frac{1}{2}\ln L.$$ Taking derivatives with respect to $u$ gives $$\frac{T'}{T} = \frac{1}{2}\cdot\frac{L'}{L} = \frac{kL}{2L} = \frac{k}{2}.$$