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

Answer 1

(b) −n²x

\[x = a \cos nt - b \sin nt\]

\[\text { Differentiating w . r . t . t, we get }\]

\[\frac{d x}{d t} = - an \sin nt - bn \cos nt\]

\[\text { Differentiating again w . r . t . t, we get }\]

\[\frac{d^2 x}{d t^2} = - a n^2 \cos nt + b n^2 \sin nt\]

\[ = - n^2 \left( a\cos nt - b \sin nt \right)\]

\[ = - n^2 x\]