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The period of the free oscillations of the system shown here if mass m1 is pulled down a little and force constant of spring is k and masses of fixed pulleys are negligible is

The period of the free oscillations of the system shown here if mass m1 is pulled down a little and force constant of spring is k and masses of fixed pulleys are negligible is

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Grade:12th pass

1 Answers

Arun
25750 Points
4 years ago
Dear Neha
Call the tension in the cord be T. Net force up on the fixed pulley is 
F1 = T - M1 g = M1 a = M1 y'' 

Net force F2 up on the movable pulley of mass M2 is 
F2 = T + T - M2 g - k h = M2 h'' 

where h is the difference in length of the spring from its rest length. If the spring length increases by h, the length of the cord from the movable pulley to the upper support decreases by h, and the length of the cord from the movable pulley to the fixed pulley also decreases by h. If the total length of the cord is fixed, the height of mass M1 above the lower support must decrease by 2h, so that 
dy/dt = -2 dh/dt --> d^2y/dt^2 = -2 d^2h/dt^2 
F1 = T - M1 g = M1 y'' = M1 (-2 h'') --> T = M1 (-2 h'') + M1 g 
F2 = 2 T - M2 g - k h = M2 h'' --> T = (M2 h'' + k h + M2 g) / 2 
M1 (-2 h'') + M1 g = T = (M2 h'' + k h + M2 g) / 2 
(M2 / 2 + 2 M1) h'' + (k/2) h = (M1 g - M2 g/2) 

Since all you want are the natural frequencies, assume a solution of the form h = A exp( 2pi t / T) (or else of the form A cos(2pi t/T) if you're bothered by complex exponentials), plug it into the homogeneous linear differential equation 
(M2 / 2 + 2 M1) h'' + (k/2) h = 0 

and solve for T: 
- (M2 / 2 + 2 M1) (2pi / T)^2 h + (k/2) h = 0 
--> (2pi / T)^2 = (k/2) / (M2 / 2 + 2 M1) 
T = 2pi sqrt( (M2 / 2 + 2 M1) / (k/2) ) 
hence option C si correct
 

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