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 w.r.t.t, we get
$\frac{\text{dx}}{\text{dt}}=-\text{an}\sin\text{nt}-\text{bn}\cos\text{nt}$
Differentiating 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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