I came across this question in Pure Mathematics 2 in The Binomial Theorem section.
The first part is simple and goes like this:
If a clock with a seconds pendulum registers x seconds too few per day, what is the time of one beat of the pendulum?
Since there are $24\times60\times60$ seconds in a day, and the pendulum beats x less than the number of seconds in a day it follows that the time for one beat would be$$\frac{Total- Time}{Number-of- Beats} =\frac{86400}{86400-x}$$ The second part is as follows:
One beat of a seconds pendulum takes $\pi(\frac{L}{g})^\frac{1}{2}$ seconds, where L is the length of the pendulum and g is a constant . If the length of the pendulum increases by $0.04$% owing to expansion, calculate the number of seconds it will have failed to register in a day.
I took this to mean:
$$\frac{86400}{86400-x} = \pi(\frac{L}{g})^\frac{1}{2}$$
Making x the subject: $$x = 86400(1 - \frac{1}{\pi(\frac{L}{g})^\frac{1}{2}})$$
And by increasing L by $0.04$% we should be able to determine the new value of x $\rightarrow$x'
$$x' = 86400(1 - \frac{1}{\pi(\frac{L\times1.0004}{g})^\frac{1}{2}}$$
In terms of x: $$x' = \frac{\pi-\frac{1}{(\frac{L}{g})^\frac{1}{2}}}{\pi-\frac{1}{(\frac{L\times1.0004}{g})^\frac{1}{2}}}x$$
This is as far as I got but apparently the answer is 17. Is this really the answer? And if so, how? And where does Binomial Theorem even apply here?
I'm guessing you are expected to use the Binomial Approximation here, ie, for $|x|<<1$, $(1+x)^n=1+nx$. Essentially, the terms from ${n \choose 2}x^2$ and onward get neglected due to the fact that $|x|<<1$ so $x^2<<|x|$.
Now notice that one beat of our regular pendulum was taking 1 second,
$\frac{86400}{86400}=\pi\sqrt\frac{L}{g}$.
The time for one beat of faulty clock would be $\frac{86400}{86400-x}=\pi\sqrt\frac{L*1.0004}{g}$.
If we took the ratio of both, we would get
$\frac{86400}{86400-x}=\sqrt\frac{L*1.0004}{L}=\sqrt{1.0004}$ here is where we use binomial approximation to say $(1.0004)^\frac{1}{2}=1.0002$ (Well using a calculator does show that the square root is $1.00019998$, so this is a very good approximation).
So we just need to solve $\frac{86400}{86400-x}=1.0002$ which gives $x=17.276$ seconds.