If x=a nt-b nt then d^2x dt^2 is: - Vidyadip

MCQ

If $\text{x}=\text{a}\cos\ \text{nt}-\text{b}\sin\ \text{nt}$ then $\frac{\text{d}^2\text{x}}{\text{dt}^2}$ is:

A
$n^2x$
✓
$-n^2x$
C
$-nx$
D
$nx$

✓

Answer

Correct option: B.

$-n^2x$

Here
$\text{x}=\text{a}\cos\text{nt}-\text{b}\sin\text{nt}$
Differentiating $\text{w.r.t. t}$, we get
$\frac{\text{dx}}{\text{dt}}=-\text{an}\sin\text{nt}-\text{bn}\cos\text{nt}$
Differentiating $\text{ w.r.t. t}$, we get
$\frac{\text{d}^2\text{x}}{\text{dt}^2}=-\text{an}^2\cos\text{nt}+\text{bn}^2\sin\text{nt}$
$=-\text{n}^2\text{a}\cos\text{nt}-\text{b}\sin\text{nt}$
$=-\text{n}^2\text{x}$

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