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Question
If x = a cos nt − b sin nt, then \[\frac{d^2 x}{d t^2}\] is
Options
n2 x
−n2 x
−nx
nx
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Solution
(b) −n2x
\[x = a \cos nt - b \sin nt\]
\[\text { Differentiating w . r . t . t, we get }\]
\[\frac{d x}{d t} = - an \sin nt - bn \cos nt\]
\[\text { Differentiating again w . r . t . t, we get }\]
\[\frac{d^2 x}{d t^2} = - a n^2 \cos nt + b n^2 \sin nt\]
\[ = - n^2 \left( a\cos nt - b \sin nt \right)\]
\[ = - n^2 x\]
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