MCQ
If $x = A\cos 4t + B\sin 4t$, then ${{{d^2}x} \over {d{t^2}}} = $
- ✓$-16x$
- B$16 x$
- C$x$
- D$-x$
✓
Answer
Correct option: A.
$-16x$
a
(a) $x = A\cos 4t + B\sin 4t$
(a) $x = A\cos 4t + B\sin 4t$
Differentiate w.r.t. $t,$
$\frac{{dx}}{{dt}} = - 4A\sin 4t + 4B\cos 4t$
Again, differentiate w.r.t. $t,$
$\frac{{{d^2}x}}{{d{t^2}}} = - 16A\cos 4t - 16B\sin 4t$
$\frac{{{d^2}x}}{{d{t^2}}} = - 16[A\cos 4t + B\sin 4t]$.
Hence, $\frac{{{d^2}x}}{{d{t^2}}} = - 16x$.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
$\int\limits_0^{\frac{\pi }{2}} {\,\,\frac{{dx}}{{1\,\, + \,\,{a^2}{{\sin }^2}x}}} $ has the value :
→Choose the correct answers from the given four options:The function $\text{f(x)}=\text{e}^{|\text{x}|}$ is:
→Let $X$ denote the number of times heads occur in $n$ tosses of a fair coin. If $P(X = 4), P(X = 5)$ and $P(X = 6)$ are in $AP,$ the value of $n$ is:
→Consider a square of side $2\ cm$ . It is cut from one of its corner as shown in adjacent figure. Maximum value of sum of the perimeters of two plane figures thus formed is
→Which of the following represents coinitial vector:
Solution of the differential equation
→$x = 1 + xy\frac{{dy}}{{dx}} + \frac{{{{\left( {xy} \right)}^2}}}{{2!}}{\left( {\frac{{dy}}{{dx}}} \right)^2} + \frac{{{{\left( {xy} \right)}^3}}}{{3!}}{\left( {\frac{{dy}}{{dx}}} \right)^3} + ......$ is
If $y = 2x + {\cot ^{ - 1}}\,x + \log \left( {\sqrt {1 + {x^2}} - x} \right),$ then $y$
→Let $\vec{u}=2 \hat{i}-\hat{j}+\hat{k}, \vec{v}=-3 \hat{j}+2 \hat{k}$ be vectors in $R^3$ and $\vec{w}$ be a unit vector in the $X Y$-plane. Then, the
→Choose the correct answer from the given four options : At $\text{x}=\frac{5\pi}{6},\text{ f(x)}=2\sin3\text{x}+3\cos\text{x}$ is $6 :$
→Area of the region bounded by the curve $y=\tan x$, line $x=\frac{\pi}{4}$ and the $x$-axis is
→