A mass m = 50 g is dropped on a vertical spring of spring constant 500 N nfl 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 principasl axis coinciding with the lilne of motion of the mss, 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

Explanation

x: maximum compression of spring
Applying energy conservation (1) and (2)
`mg(h+x)=(1)/(2)kx^2`
`(50)/(1000)x10(0.1+x)=(1)/(2)xx500x^2`
`500x^2-x-0.1=0`
`x=(1+-sqrt(1+4xx500xx0.1))/(2xx500)=0.015m=1.5cm`
The block is in equilibrium when `mg=kx_0`
`(50)/(1000)xx10=500x_0impliesx_0=10^-3m=0.1cm`
Amplitude of oscillation `=x-x_0=1.5-0.1=1.4cm`
O: equilibrium position `PO=30.1cm,PB=31.5cm`
image of A:
`(1)/(v_A)+(1)/(-28.7)=(1)/(-12)`
`v_A=-20.6cm`
Image of B:
`(1)/(v_B)+(1)/(-31.5)=(1)/(-12)`
`v_B=-19.4cm`
The length in which image oscillates
`=20.6-19.4=1.2cm`