Question

The differential equation of $SHM$ for a particle is given by $a\frac{d^{2}x}{d{t}^2}+bx=0$. The ratio of the maximum acceleration to the maximum velocity of the particle is

• A. $\sqrt{\frac{b}{a}}$
  
• B. $\frac{a}{b}$
  
• C. $\frac{b}{a}$
  
• D. $\sqrt{\frac{a}{b}}$
  

Answer 1

The correct option is A $\sqrt{\frac{b}{a}}$


Given:
$a\frac{d^{2}x}{d{t}^2}+bx=0\text{........ (1)}$

General solution of $S.H.M.$ $x=A\sin (\omega t+\varphi )\dots \dots \dots ..\left(2\right)$ Differentiate $\left(2\right)$ equation with respect to ' $t$ '
velocity ,$\frac{dx}{dt}=A\omega \cos (\omega t+\varphi )$
Again differentiating

acceleration $\frac{d^{2}x}{d{t}^2}=-A{\omega }^2\sin (\omega t+\varphi )=-{\omega }^{2}x$

$\frac{d^{2}x}{d{t}^2}+{\omega }^{2}x=0$......... $\left(3\right)$

Dividing with ' $a$ ' in equation $\left(1\right)$

$\frac{d^{2}x}{d{t}^2}+\frac{b}{a}x=0$....... $\left(4\right)$
comparing equations $\left(3\right)$ and $\left(4\right)$

${\omega }^2=\frac{b}{a}\Rightarrow \omega =\sqrt{\frac{b}{a}}$

maximum acceleration $a_{max}={\omega }^{2}A\text{and}$
maximum velocity $v_{max}=\omega A$

$\therefore$ $\frac{a_{max}}{v_{max}}=\omega$

or, $\frac{a_{max}}{v_{max}}=\sqrt{\frac{b}{a}}$

Hence , option $\left(D\right)$ is correct.