A classical mechanics problem by Milun Moghe

Classical Mechanics Level 4

In a certain hypothetical space there exists a plane SCOD in which a point source exists at S which radiates light radially only in the given plane. The long line segments extended even after OC and OD have equal masses m=3^0.5 and are hinged about O. OC and OD uniformly distributed masses and are perfect ideal light reflectors. Internal bisector of the two lines is shown by an imaginary dotted line.An arc EGF made of some compressible material and is a perfect absorber of light .

Any light radiation falling on the arc at any angle is absorbed completely. Out of all the light rays emitted from the source of power P=1watt the initial path of two rays are given.

At t=0 the source is switched ON. By some mechanism inside the part EOFG is an energy converter and oscillator which makes the lines oscillate with a very small amplitude with some instantaneous frequency (through an axis perpendicular to the given plane passing through O at time t=0) given by

f = k ( P a ) 3 m 4 d f=k(P_{a})^{3}m^{4}d

Here K is constant P a = P o w e r ( a b s o r b e d ) P_{a}=Power(absorbed) ,d is the distance of the fixed source from the fixed imaginary axis (internal bisector of the lines of dotted lines)

The lines even though oscillate do not affect the circular shape of the arc only angle is reduced or increased

S M = 5 u n i t s SM=5units , B O = b = 100 u n i t s BO=b=100units M is a foot of pipendicular dropped on the axis from the source

A O = a = 100 3 0.5 u n i t s AO=a=\frac{100}{3^{0.5}}units E O = G O = F O = 50 u n i t s EO=GO=FO=50units t a n θ 2 = 1 3 0.5 tan\theta_{2}=\frac{1}{3^{0.5}} t a n θ 1 = 3 0.5 tan\theta_{1}=3^{0.5}

initially α = π 20 \alpha=\frac{\pi}{20}

Find the instantaneous frequency at t=7 seconds... the value of k=64

Neglect any other radiations and assume reflecting properties of light same as that when reflector is at rest , at all instants of time.


The answer is 45.

Milun Moghe by Milun Moghe , 25, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Milun Moghe
Feb 11, 2014

The problem is a very illustrative example of useful technique . I have also given some unnecessary information of initial angle alpha just for making the question answer look complicated.A pragmatic solution to the problem involves thinking little different. A normal student would think that given two rays how could the power absorbed be calculated. This technique is a very powerful technique used to calculate with ease shortest distance during number of reflections in a conical or such triangular reflectors. In the picture is a diagram of how we can represent multiple reflections by one single ray is shown. Just keep on flipping the surface about A1 then A2 again A1 and so on . By this technique we can easily find the shortest distance of the ray from O. If the shortest distance is less than the radius of the absorber then light is absorbed otherwise not the shortest distance OX can be found out as O X = d s i n θ 1 OX=dsin\theta_{1} From the information given in the question we can see that a s i n θ 1 = b s i n θ 2 = r = E O = 50 asin\theta_{1}=bsin\theta_{2}=r=EO=50 So all rays within the smaller angle CSD will be absorbed and rest will not. fortunately due to geometry the ∠ C S D = θ 1 + θ 2 = π / 2 \angle CSD=\theta_{1}+\theta_{2}=\pi/2 which does not depend on alpha P a = P π 2 2 π = 0.25 w a t t P_{a}=P\frac{\frac{\pi}{2}}{2\pi}=0.25watt

So due to oscillations of the lines power does not change thus frequency remains constant = 45 h e r t z \boxed{45hertz}

Unfortunately i dont know how to upload images can any one help me out at this

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...