Question

If $x = a\cos nt - b\sin nt$, then $\dfrac{{{d^2}x}}{{d{t^2}}}$
A. \[{n^2}x\]
B. $ - {n^2}x$
C. $ - nx$
D. $nx$

Answer 1

Hint:
First, differentiate $x = a\cos nt - b\sin nt$ with respect to $t$ where, $\dfrac{d}{{dt}}\left( {\cos nt} \right) = - n\sin t$ and $\dfrac{d}{{dt}}\left( {\sin nt} \right) = n\cos t$. Next, differentiate again to find the value of $\dfrac{{{d^2}x}}{{d{t^2}}}$. At last, substitute the value of $x = a\cos nt - b\sin nt$ to simplify the equation.

Complete step by step solution:
We are given that $x = a\cos nt - b\sin nt$
We have to find the value of $\dfrac{{{d^2}x}}{{d{t^2}}}$
We will differentiate the given equation with respect to $t$
We know that $\dfrac{d}{{dt}}\left( {\cos nt} \right) = - n\sin t$ and $\dfrac{d}{{dt}}\left( {\sin nt} \right) = n\cos t$
Then, the differentiation of $x = a\cos nt - b\sin nt$
$\dfrac{{dx}}{{dt}} = - na\sin nt - nb\cos nt$
We will again differentiate the above equation with respect to $t$.
$\dfrac{{{d^2}x}}{{d{t^2}}} = - {n^2}a\cos nt + {n^2}b\sin nt$
We will ${n^2}$ common from RHS
$\dfrac{{{d^2}x}}{{d{t^2}}} = - {n^2}\left( {a\cos nt - b\sin nt} \right)$
And from the given equation, the value of $\left( {a\cos nt - b\sin nt} \right)$ is $x$
Therefore,
$\dfrac{{{d^2}x}}{{d{t^2}}} = - {n^2}x$

Hence, option B is correct.

Note:
Students must remember the formulas of differentiation to do these types of questions. Also, the differentiation of $af\left( x \right)$ where $a$ is constant, then the differentiation is $af'\left( x \right)$ and differentiation of $f\left( x \right) \pm g\left( x \right)$ is $f'\left( x \right) \pm g'\left( x \right)$.