Question

Given a differential equation of SHM is :

$a\frac{d^{2}x}{d{t}^2}+bx=0$

The ratio of the maximum acceleration to the maximum velocity of the particle is:

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

Answer 1

Answer: (D)

$\sqrt{b/a}$

The given differential equation of SHM is:
$a\frac{d^{2}x}{d{t}^2}+bx=0$
Dividing whole equation by $a$, we have:

$\frac{d^{2}x}{d{t}^2}+\frac{b}{a}x=0$
Now, compare it with the standard differential equation of SHM i.e. $\frac{d^{2}x}{d{t}^2}+{\omega }^{2}x=0$, we have
${\omega }^2=\frac{b}{a}$
We know that:

$a_{max}={\omega }^{2}A$ and $v_{max}=\omega A$
$\Rightarrow \frac{a_{max}}{v_{max}}=\omega$

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