## PROBLEMS

modernphysics
Posts: 33
Joined: Fri Jun 11, 2010 10:50 am

### PROBLEMS

1. A detector is used to count the number of $$\gamma\$$ rays emitted by a radio active source. If the number of counts recorded in exactly 20 secs. is 10000, the error in the counting rate of per second is what?
2. A point particle is moving in the (X,Y) plane on a trajectory given in polar coordinates by the equation $$r^2$$-2rsin($$\theta\$$+$$\pi\$$/4)-3=0.
Then the trajectory of the particle is what?

THANK YOU....

michael
Posts: 50
Joined: Fri Sep 05, 2008 7:21 am

### Re: PROBLEMS

1. A detector is used to count the number of \gamma\ rays emitted by a radio active source. If the number of counts recorded in exactly 20 secs. is 10000, the error in the counting rate of per second is what?

AVERAGE counts per second = 10000/20 = 500. The error (standard deviation) in the number of counts is the square root of the number of counts (according to poisson statistics which apply to systems such as this where the average rate is constant and events are independent from one-another) and therefore the uncertainty in 500 counts per second is about 22

2. A point particle is moving in the (X,Y) plane on a trajectory given in polar coordinates by the equation $$r^2-2rsin(\theta+\pi/4)-3=0$$.
Then the trajectory of the particle is what?

Its not clear to me exactly what you mean by trajectory here, since as far as I am concerned, the equation written above defines the trajectory. If you want the equation which defines the trajectory in terms of X and Y, then you can use this method.
First note how to change variables:
$$r^2 = X^2 + Y^2$$
$$sin(\theta) = Y/\sqrt{X^2 + Y^2}$$
$$cos(\theta) = X/\sqrt{X^2 + Y^2}$$
The idea is to just sub these in to replace r and theta by X and Y, but it looks clear that the sine part will be ugly, so split it first into stuff using the sin rule [sin(A+B) =sinA cosB - sinB cosA] which we can use more easily with our transformations between co-ordinate systems:
$$r^2-2rsin(\theta) cos(\pi/4) + 2rcos(\theta)sin(\pi/4) -3=0$$
which is
$$r^2-2rsin(\theta) \sqrt{3}/2 + rcos(\theta) -3=0$$
Now substituting in for X and Y:
$$X^2 + Y^2 -\sqrt{3} Y + X -3=0$$
In principal you could use the quadratic formula to write this as X = f(Y) for some function f, but I am pretty sure it will be very ugly, so the last equation I wrote is probably what they want as the final solution, assuming i didnt make any mistakes.

Hope that helped

modernphysics
Posts: 33
Joined: Fri Jun 11, 2010 10:50 am

### Re: PROBLEMS

michael wrote:1. A detector is used to count the number of \gamma\ rays emitted by a radio active source. If the number of counts recorded in exactly 20 secs. is 10000, the error in the counting rate of per second is what?

AVERAGE counts per second = 10000/20 = 500. The error (standard deviation) in the number of counts is the square root of the number of counts (according to poisson statistics which apply to systems such as this where the average rate is constant and events are independent from one-another) and therefore the uncertainty in 500 counts per second is about 22

2. A point particle is moving in the (X,Y) plane on a trajectory given in polar coordinates by the equation $$r^2-2rsin(\theta+\pi/4)-3=0$$.
Then the trajectory of the particle is what?

Its not clear to me exactly what you mean by trajectory here, since as far as I am concerned, the equation written above defines the trajectory. If you want the equation which defines the trajectory in terms of X and Y, then you can use this method.
First note how to change variables:
$$r^2 = X^2 + Y^2$$
$$sin(\theta) = Y/\sqrt{X^2 + Y^2}$$
$$cos(\theta) = X/\sqrt{X^2 + Y^2}$$
The idea is to just sub these in to replace r and theta by X and Y, but it looks clear that the sine part will be ugly, so split it first into stuff using the sin rule [sin(A+B) =sinA cosB - sinB cosA] which we can use more easily with our transformations between co-ordinate systems:
$$r^2-2rsin(\theta) cos(\pi/4) + 2rcos(\theta)sin(\pi/4) -3=0$$
which is
$$r^2-2rsin(\theta) \sqrt{3}/2 + rcos(\theta) -3=0$$
Now substituting in for X and Y:
$$X^2 + Y^2 -\sqrt{3} Y + X -3=0$$
In principal you could use the quadratic formula to write this as X = f(Y) for some function f, but I am pretty sure it will be very ugly, so the last equation I wrote is probably what they want as the final solution, assuming i didnt make any mistakes.

Hope that helped
Thank You My Dear Friend,
The first solution is right methinks.
But in the second one you put the formula and values wrongly and end of getting a wrong equation all together to the
given parametric equation. Thats it. Otherwise you are done in CLEARING MY DOUBTS.
Error:
$$1. sin(\alpha+\beta)=sin(\alpha)cos(\beta)+cos(\alpha)sin(\beta)$$
$$2. sin(\pi/4)=cos(\pi/4)=1/\sqrt2$$

And after doing as you have done i got the equation as
$$X^2+Y^2-\sqrt{2}X-\sqrt{2}Y-3=0$$

The equation of a "CIRCLE". That was what i meant by what its trajectory is..
So thanks... THANK U..You showed me the path.

michael
Posts: 50
Joined: Fri Sep 05, 2008 7:21 am

### Re: PROBLEMS

Your corrections look right to me (i looked up the relations this time haha). Sorry for the errors - its quite worrying for me that I have forgotten these things, I am clearly out of practice! Glad I was able to help with the method at least.

babbar.ankit
Posts: 5
Joined: Sat May 29, 2010 11:56 am

### Re: PROBLEMS

I believe the right way for the first part would be rather==>
Err=√N;Err(/s)=√(10000)/20=5; which also signify that the counting for longer times lead to lesser error...
(The soln earlier proposed was N(/s)=500;=>Err(/s)=√(500)=22)
Plz correct me if i am wrong