Question

On applying a force F the point 'P' is displaced vertically downwards by y units from equilibrium position. Find the force F in terms of the force constant k of the spring and displacement y, for the cases (A) and (B), as shown in figure.

• A. Case A : F = ky Case B : F = ky
• B. Case A : F = 2ky Case B : F = ky
• C. Case A : F = ky Case B : F = 4ky
• D. Case A : F = ky Case B : F = 2ky

Answer 1

The correct option is C

Case A : F = ky

Case B : F = 4ky

Case (A)

At point P : F = T ------------------(i)

And for the equilibrium of pulley, 2T = k$x_0$ ----------(ii)

But as due to shift of point P by y, the spring stretches by ($\frac{y}{2}$),

so $F_s=k\left(\frac{y}{2}\right)$

so substituting $F_s$ from Eqn. (iii) in Eqn. (ii)

and then T from Eqn. (ii) in Eqn. (i), we get F = ($\frac{k}{4}$)y ...(A)

Case (B) As tension in massless string and spring will be same,T = kx ...(i)

For pulley : F = 2 Kx ...(ii)

Now if the point P shifts by y the spring will stretch by 2y (as string is inextensible)

F=2k(2y) ...(iii)

F = (4k) y ...(B)