Physics - SHM Question with Solution | TestHub

Let, distance moved by A and B in any time interval as measured from natural length of spring are xA and xB respectively. Let,$l$is the natural length of spring, then elongation in spring at any instant is

$\text{x}=\left({\text{x}}_{\mathrm{A}}-{\text{x}}_{\mathrm{B}}-l\right)$

$\text{U}=\frac{{\text{kx}}^2}{2}$

$\frac{\text{dU}}{\text{dt}}=\text{P}=\frac{\text{k}}{2}\times 2\text{x}\frac{\text{dx}}{\text{dt}}$

$\Rightarrow 20=\text{kx }\left[{\text{v}}_{\text{A}}-{\text{v}}_{\text{B}}\right]$

$\Rightarrow \text{kx}=\frac{20}{4-2}=1\mathrm{0 }\text{N}$