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I am given $\omega\left ( k \right )=\left ( \frac{C+H}{m}\pm \frac{1}{m}\sqrt{C^{2}+H^{2}+2 CH Cos\left ( k a \right )} \right )^{1/2}$.

It is mentioned that this reduces to

$\omega\left ( k \right )=\sqrt{\frac{CH}{2m\left ( C+H \right )}}a\left | k \right |$ when $\lambda \rightarrow \infty$ for $\lambda = \frac{2 \pi}{k}$.

After 5 hours of frustrating attempt, I am unable to progress further.

Attempt:

Note that in the event that $\lambda \rightarrow \infty, k\rightarrow 0$.

Thence, it can be expected that $ka \approx 0$ and so $Cos\left ( ka \right )\approx 1$.

It is useful to recall that for small angle $\left ( ka \right ), Cos\left ( ka \right )\approx 1-\frac{\left ( ka \right )^{2}}{2}$.

Expanding this we obtain at

$\left ( \frac{C+H}{m}-\frac{1}{m}\sqrt{C^{2}+H^{2}+2CH} \right ) =0$

which isn't what I am looking for.

Any help is appreciated.

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Hint

As you wrote, you need to use $$\cos(ka)=1-\frac{a^2 k^2}{2}+O\left(a^4\right)$$ which makes $$C^{2}+H^{2}+2 CH \cos\left ( k a \right )=(C+H)^2-CH a^2k^2+O\left(a^4\right)$$ $$\sqrt{C^{2}+H^{2}+2 CH \cos\left ( k a \right )}\sim(C+H)\sqrt{1-\frac{CHa^2k ^2}{(C+H)^2}}\sim(C+H)\left(1-\frac{CHa^2k ^2}{2(C+H)^2} \right)$$

Just continue.

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  • $\begingroup$ But if k is small, why isn't it valid to neglect $(ka)^{2}$? $\endgroup$ – Mathematicing Oct 31 '17 at 6:42
  • $\begingroup$ @Mathematicing. Just because you want an approximation ! Otherwise, as you did, you would get $0$ just as you did. Just try with a calculator using, say, $C=H=1$ and $ka=\frac \pi {24}$ for which $\cos(ka)=\frac{1}{2} \sqrt{2+\sqrt{2+\sqrt{3}}}$ $\endgroup$ – Claude Leibovici Oct 31 '17 at 6:49
  • $\begingroup$ I'm also having a bit of trouble understanding to second approximation equality. Is this a result of purely algebraic manipulation of approximation? $\endgroup$ – Mathematicing Oct 31 '17 at 6:58
  • $\begingroup$ @Mathematicing. You need to continue the process $$\frac{C+H}{m}-\frac{1}{m}\sqrt{C^{2}+H^{2}+2 CH \cos\left ( k a \right )}=\frac {CHa^2k^2}{2m(C+H)}$$ and you need the square root of that. All of that is based on manipulations using Taylor series or binomial expansion. $\endgroup$ – Claude Leibovici Oct 31 '17 at 7:06
  • $\begingroup$ I'm afraid I am unable to arrive at the final approximation in your last line. $\endgroup$ – Mathematicing Oct 31 '17 at 7:22

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