Question

Two blocks A and B, each of mass m are connected by means of a pulley-spring system on a smooth inclined plane of inclination $\theta$ as shown in the figure. All the pulleys and spring are ideal. Now, B is slightly displaced from its equilibrium position. It starts to oscillate. Time period of oscillation of B will be (Take m = 4 kg, K = 5 N/m, $\pi$ = 3.14)

• A. 3.14 s
• B. 6.28 s
• C. 4.28 s
• D. 5.14 s

Answer 1

The correct option is B 6.28 s

Let elongation of spring be x0 in equilibrium. Then,

$2T+mgsin\theta =2k{x}_0\dots \left(i\right)$

and $T=mg\dots \left(ii\right)$

Let Block B is displaced by x down the inclination

F.B.D. of B


$-m{a}_B=2K(x_0+2x)-2T^{\prime }-mgsin\theta \dots \left(iii\right)$

F.B.D. of A

$mg-T^{\prime }=m{a}_A$

Also, $a_A=2a_B$

$T^{\prime }=mg-2m{a}_B$

$-m{a}_3=2k{x}_0+4kx-2mg+4m{a}_3-mgsin\theta$

$-m{a}_3=4kx+m{a}_3$

$a_B=-\frac{4k}{5m}x$

$\therefore T=2\pi \sqrt{\frac{5m}{4k}}$

$T=6.28s.$