A mass m =50g is dropped on a vertical spring of spring constant 500 N/m from a height h =10 cm as shown in figure. The mass sticks to the spring and executed 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 oscillated
mg = kx
Now mean position at 30 + 0.1 = 30.1 cm from mirror
Mg (n + y) = 1/2 ky2
u1 = – 31.5 cm, – (31.5 – 30.1) = – 1.4 cm
f = – 12 cm f = – 12 cm, u2 = 30.1 – 1.4 = – 28.7 cm
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