Two identical particles A and B, each of mass m, are interconnected by a spring of stiffness k. If the particle B experiences a force F the elongation of the relative spring is x, the relative acceleration between the particles is equal to:

Question

Two identical particles A and B, each of mass m, are interconnected by a spring of stiffness k. If the particle B experiences a force F the elongation of the relative spring is x, the relative acceleration between the particles is equal to: (A) F / 2 m (B) ( F - kx / m ) (C) (F-2 k x/m) (D) (k x/m)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/434fa3c1885719d018781743c17ce99a-.png" /> Equation of motion for A: T=m a ⇒ a =( T / m )=( kx / m ) for B: F - T = ma prime ⇒ a prime=( F - kx / m ) ⇒ The relative acceleration ar=|a prime-a|=(F-2 k x/m)

Q. Two identical particles $A$ and $B$ , each of mass $m$ , are interconnected by a spring of stiffness $k$ . If the particle B experiences a force $F &$ the elongation of the relative spring is $x$ , the relative acceleration between the particles is equal to:

$F /2 m$

$\frac{F - k x}{m}$

$\frac{F - 2 k x}{m}$

$\frac{k x}{m}$

Equation of motion for $A : T = ma$
$\Rightarrow a = \frac{T}{m} = \frac{k x}{m}$ for $B : F - T = m a^{'}$
$\Rightarrow a^{'} = \frac{F - k x}{m}$
$\Rightarrow$ The relative acceleration
$a_{r} = ∣ a^{'} - a ∣ = \frac{F - 2 k x}{m}$

Q. Two identical particles A and B, each of mass m, are interconnected by a spring of stiffness k. If the particle B experiences a force F& the elongation of the relative spring is x, the relative acceleration between the particles is equal to:

F/2m

mF−kx​

mF−2kx​

mkx​