Question

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 $$T=2 \pi \sqrt{\frac{L}{g}}$$ 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 In symbols, with \(u\) being temperature and \(k\) the proportionality constant, $$\frac{d L}{d u}=k L$$ Assuming this to be the case, show that the rate at which the period changes with respect to temperature is \(k T / 2\)

Short answer

The rate at which the period changes with respect to temperature is \(kT/2\).

Step-by-step solution

Find the derivative of period with respect to length

We start with the given formula for the period of a pendulum, i.e., \(T = 2\pi\sqrt{\frac{L}{g}}\) and we want to take the derivative of this with respect to \(L\) (denoted as \(dT/dL\)). By applying the chain rule, we have \(\frac{dT}{dL} = 2\pi \cdot \frac{1}{2} \cdot (\frac{L}{g})^{-1/2} \cdot \frac{1}{g}=\frac{\pi}{\sqrt{gL^3}}\).

Apply the chain rule again

In this step, we need to find the derivative of the period with respect to temperature. Using the chain rule again, we get that \(\frac{dT}{du} = \frac{dT}{dL} \cdot \frac{dL}{du}\). By substituting the given \(\frac{dL}{du} = kL\) and the derived \(\frac{dT}{dL} = \frac{\pi}{\sqrt{gL^3}}\) into this equation, we get \(\frac{dT}{du} = kL \cdot \frac{\pi}{\sqrt{gL^3}}\).

Simplify the answer

Now we can simplify the result. We know that the period of the pendulum is given by \(T = 2\pi\sqrt{\frac{L}{g}}\). Therefore, substituting \(T = 2\pi\sqrt{\frac{L}{g}}\) into \(\frac{dT}{du} = kL \cdot \frac{\pi}{\sqrt{gL^3}}\) gives us: \(\frac{dT}{du} = kT/2\).