Physics - SHM Question with Solution | TestHub

For a particle executing linear S.H.M., the magnitude of acceleration at a distance $x$ from the mean position is given by $a={\omega }^{2}x$.

Since $0<a_1<a_2$, both positions are on the same side of the mean position. Let their distances from the mean position be $x_1$ and $x_2$ respectively, with $x_1<x_2$.

The distance between these two positions is $d=x_2-x_1$.

The velocities at these positions are given by:
$V_1^2={\omega }^2(A^2-x_1^2)$
$V_2^2={\omega }^2(A^2-x_2^2)$

Subtracting the two equations, we get:
$V_1^2-V_2^2={\omega }^2(x_2^2-x_1^2)$
$V_1^2-V_2^2={\omega }^2(x_2-x_1)(x_2+x_1)$

We know that $a_1={\omega }^2{x}_1$ and $a_2={\omega }^2{x}_2$. Adding these gives:
$a_1+a_2={\omega }^2(x_1+x_2)$

Substituting this into the velocity difference equation:
$V_1^2-V_2^2=(x_2-x_1)(a_1+a_2)$

Since $d=x_2-x_1$, we have:
$V_1^2-V_2^2=d(a_1+a_2)$

$\Rightarrow d=\frac{V_1^2-V_2^2}{a_1+a_2}$

Answer: $\frac{V_1^2-V_2^2}{a_1+a_2}$