Question:
Write the correct alternative in the following:
If $x=a \cos n t-b \sin n t$, then $\frac{d^{2} x}{d t^{2}}$ is
A. $n^{2} x$
B. $-n^{2} x$
C. $-n x$
D. $n x$
Solution:
Given:
$x=a \cos n t-b \sin n t$
$\frac{\mathrm{dx}}{\mathrm{dt}}=-\mathrm{an} \sin \mathrm{nt}-\mathrm{bn} \cos \mathrm{nt}$
$\frac{d^{2} x}{d t^{2}}=-a n^{2} \cos n t+b n^{2} \sin n t$
$=-n^{2}(a \cos n t-b \sin n t)$
$=-n^{2} x$
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