Question

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. 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.

Answer 1

The length in which the image of the mass oscillates = 1.24 cm.

Explanation:

Because of the body's weight suppose the spring is compressed by which the oscillation's mean location is,

$\begin{equation} M=50 \times 10^{-3} \mathrm{kg}\end$

$\begin{equation}g=10 \mathrm{m} / \mathrm{s}^{2}\end$

$\begin{equation}\mathrm{k}=500 \mathrm{N} / \mathrm{m}^{2}\end$

h = 10 cm = 0.1 m  

For equilibrium,      

mg = kx  

$\begin{equation}\frac{m g}{k}=x\end$    

$\begin{equation}x=\frac{50 \times 10^{-3} \times 10}{500}\end$

$\begin{equation}x=\frac{10^{-3} \times 10}{10}\end$

$\begin{equation}x=10^{-3} \mathrm{m}=0.1 \mathrm{cm}\end$

Therefore, the mean position is at 30+0.1 = 30.1 cm

Suppose optimum spring compression is δ

Since,

The principle of energy work

$\begin{equation}M g(h+\delta)-\frac{1}{2} k \delta^{2}=0\end$

$\begin{equation}M g(h+\delta)=\frac{1}{2} k \delta^{2}\end$

$\begin{equation}50 \times 10^{-3} \times 10(0.1+\delta)=\frac{1}{2} \times 500 \delta^{2}\end$

$\begin{equation}500 \times 10^{-3} \times(0.1+\delta)=250 \delta^{2}\end$

$\begin{equation}\left(0.05+500 \times 10^{-3} \delta\right)=250 \delta^{2}\end$

$\begin{equation}250 \delta^{2}-500 \times 10^{-3} \delta-0.05=0\end$

$\begin{equation}\delta=\frac{0.5 \pm \sqrt{0.25+50}}{2 \times 250}\end$

δ = 0.015 m = 1.5 cm  

For Position of B is 30+1.5 = 31.5

Vibration Amplitude = 31.5-30.1=1.4

Posion A is 30.1-1.4=28.7 cm from pole  

For A,  

u = -31.5

f = -12 cm  

$\begin{equation}\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\end$

$\begin{equation}\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\end$

$\begin{equation}\frac{1}{v}=-\frac{1}{12}-\frac{1}{-31.5}\end$

$\begin{equation}\frac{1}{v}=\frac{1}{31.5}-\frac{1}{12}\end$

$\begin{equation}\frac{1}{v}=\frac{12-31.5}{31.5 \times 12}\end$

$\begin{equation}\frac{1}{v}=\frac{-19.5}{378}\end$

$\begin{equation}v=\frac{-378}{19.5}\end$

v = -19.38 cm

For B,

u = -28.7

f = -12 cm  

$\begin{equation}\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\end$      

$\begin{equation}\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\end$

$\begin{equation}\frac{1}{v}=-\frac{1}{12}-\frac{1}{-28.7}\end$

$\begin{equation}\frac{1}{v}=\frac{1}{28.7}-\frac{1}{12}\end$

$\begin{equation}\frac{1}{v}=\frac{12-28.7}{28.7 \times 12}\end$

$\begin{equation}\frac{1}{v}=\frac{-16.7}{344.4}\end$

$\begin{equation}v=\frac{-344.4}{16.7}\end$

v = 20.62 cm

Thus, the projectile vibrates in length 20.62 – 19.38 = 1.24  cm.