Tiêu đề 03D.SHM-PHY LEVEL -VI ( 149 -165 ).pdf ✅

SHM 150 NISHITH Multimedia India (Pvt.) Ltd., JEE ADVANCED - VOL - III NISHITH Multimedia India (Pvt.) Ltd., four springs = 4 sin : F dF 0 0    4 0 0  x F dF l     0 0 4 . 0 F x dF x l    4F0 2 a x x ml      2 T ?      5. (B) The diagram in disturbed condition is:    l x x ' ......................(1) From diagram: y x x   sin 'sin   sin ' : sin x x           .............(2)  Restoring force, .sin F W l      A l g . . sin  .............(3) But F ma  ......................(4) where m A l  . ...................(5) sin sin h h l     ...............(6) find w, after solving above equations. 6. (A) Concept of physical pendulum 1 2 mgl f  I  where l  length between point of suspension and centre of mass. I = moment of inertia about A 10 4 L  l 2 3 ML I  7. (C) Let, x be the displacement of centre of mass of cylinder from mean position.  from rolling concept, elongation in spring= 2x  Restoring torque about point of contact is:  rest  2 2 R kx    4kxR but x R   2 4 rest      kR  I 3 2 . 2  mR  and 2        : ? 8. (B) As A oscillates up and down, the normal force between B and surface varies from minimum to maximum. To keep body B in contact with the surface, the minimum R  0 and R becomes minimum, when A is at topmost position. Free body diagrams are: 2 T m g m Aw   1 1 ..................(1) T m g R   2 .........................(2) and R  0..............................(3) solve for min f  ? 9. (B) Let x be the displacement of sleeve in disturbed condition along horizontal bar from centre. The free body diagram is:
NISHITH Multimedia India (Pvt.) Ltd., 151 JEE ADVANCED - VOL - III JEE MAINS - CW - VOL - I SHM NISHITH Multimedia India (Pvt.) Ltd., 2 Frestoring    kx m x    2    ma k m x    2 ? m m a x T m      [ do not confuse angular velocity of frame with angular frequency of SHM] 10. (B) Here interface acts as mean position of SHM:  time period, 1 2 2 2 T T T   1 2 2 1 2 : 2 l l T T g g         11. (A) Restoring torque :    mgr 2   I mgr    2 mgr I    and 2 2 12 ml I mr   ; 2 2 2 4 l r R   2 T ?      12. (C) Let the diagram in disturbed position is as shown.  Restoring torque about ‘O’ is sin cos cos 1 2     2 rest l     mg k x l k x l          sin cos  1 2  2 mgl       kxl k k k  1(     small ; sin ; cos 1;     x l ) 2 2 mgl kl          ;  I. 2 . 3 ml   2         ? 13. (A) Rolling can be considered as pure rotation about point of contact, P: Now it is compound pendulum oscillating about ‘P’. 2 2 2 2 ; where 2 I T I mR mR mR mgd       . d R  / 2   T ? 14. (A) Let ‘S’ be the surface tension of the soap film. In equilibrium position: F F upward downward  2 tan 2 2 l S mg                 4 tan / 2 Sl   m g    ..............(1)