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I have a question about the term $\frac{1}{\omega}$ in the following equation (see eq. (1.6) below)

enter image description here We have the general solution $x(t)=a\cos\omega t+b\sin\omega t$. I would think that we're substituting $b=\omega x’(0)$, so why is our initial-value solution not the following: $$ x(t)=x(0)\cos\omega t+\omega x’(0)\sin\omega t. $$

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  • $\begingroup$ Hint, take Eq 1.4 and from it calculate $x'(t)$, then evaluate it at $t=0$. $\endgroup$ – The Photon May 5 '17 at 22:53
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If you differentiate, you bring out a factor of $\omega$. In order to counteract that factor, you multiply by $x'(0)$ by $1/\omega$.

A more physical justification is to use a dimensional analysis. $\sin \omega t$ is dimensionless, $b$ has units of length, and we're solving for $b$ in terms of $x'(0)$, which has units of length/time. In order to return the units to length, we must multiply by a factor with units of time, which is the same as dividing by a factor of frequency.

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