Q77 of 123 Page 412

A mass m = 50 g is dropped on a vertical spring of spring constant 500 N m-1 from a height h = 10 cm as shown in figure (18-E14). The mass sticks to the spring and executes simple harmonic oscillations after that. A concave mirror of focal length 12 cm facing the mass is fixed with its principal axis coinciding with the line of motion of the mass, its pole being at a distance of 30 cm from the free end of the spring. Find the length in which the image of the mass oscillates.


mass of the object, m= 50g

Spring constant of the spring, k= 500N/m


Height of mass from the spring, h= 10cm


Focal length of the mirror, f= 12cm


Distance between the pole and the free end of spring = 30cm



The spring execute SHM when mass falls on it


At equilibrium position, weight of mass is equal to force applied by the spring





Therefore, mean position of SHM = 30+ 0.1= 30.1 cm (from pole of the mirror)


Let the maximum compression =


Using the work energy principle,





From the figure shown,


Position of point B= 30+1.5=31.5 cm (from the pole of the mirror)


Therefore, amplitude of vibration of SHM= 31.5-30.1= 1.4 cm


Position of the point A from the pole of the mirror= 30.1 – 1.4= 28.7 cm


For point A,


Object distance,


Using lens formula,





For point A,


Object distance,


Using lens formula,





the image vibrates in length (20.62 - 19.38) = 1.24 cm


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Two concave mirrors of equal radii of curvature R are fixed on a stand facing opposite directions. The whole system has a mass m and is kept on a frictionless horizontal table (figure 18-E15).


Two blocks A and B, each of mass m, are placed on the two sides of the stand. At t = 0, the separation between A and the mirrors is 2R and also the separation between B and the mirrors is 2R. The block B moves towards the mirror at a speed v. All collisions which take place are elastic. Taking the original position of the mirrors-stand system to be x = 0 and X-axis along AB, find the position of the images of A and B at t =



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