Question

A particle executing SHM has velocities $u$ and $v$ and accelerations $a$ and $b$ at two of its positions. Find the shortest distance between these two positions.

• A. $\frac{u^2-v^2}{a+b}$
• B. $\frac{v^2+u^2}{a-b}$
• C. $\frac{v^2+u^2}{a+b}$
• D. $\frac{v^2-u^2}{a-b}$

Answer 1

The correct option is A $\frac{u^2-v^2}{a+b}$

Let $x_1$ and $x_2$ be the distances of the two positions from mean position.
Then with usual notations,
$u^2={\omega }^2(A^2-x_1^2)$ ... (i)
$v^2={\omega }^2(A^2-x_2^2)$ ... (ii)
$a={\omega }^2{x}_1$ ... (iii)
$b={\omega }^2{x}_2$ ... (iv)

Subtracting Eq. (ii) from Eq. (i),
$u^2-v^2={\omega }^2(x_2^2-x_1^2)$ ... (v)
Adding Eqs. (iii) and (iv),
$a+b={\omega }^2(x_1+x_2)$ ... (vi)

Since we want the shortest distance, we are looking for $x_2-x_1$.
Dividing Eq. (v) by Eq. (vi)
$\frac{u^2-v^2}{a+b}=x_2-x_1$