Waves and Vibrations Questions

[محمد عمر]

AA All,

Here is a topic for all the Waves and Vibrations Questions. Add any questions concerning Waves and Vibrations into this post.

Best regards,
Mohammed Omar

[Δђ๓єđ Иαβเℓ]

ياريت الناس تقرا التوبيك ده قبل ما تحط اي اسئلة وأي سؤال يخالف اللي ذكره البشمهندس محمد سيتم حذفه مباشرة دون الرجوع للعضو

http://www.h3sonline.com/forums/showthread.php?t=12268

[diaago]

ana mesh fahem ezay el 7al bta3 mas2la rakam 3 fe as2elet el wave fe sheet doc 3esmat
7atta ana baset 3ala el 7al elly fe malazem bashmohandes el banna bs brdo mafhmtsh ya reet 7ad yewd7ly

[محمد عمر]

I could answer questions directed to me. Ask Eng. El Banna for questions concerning him.


ana mesh fahem ezay el 7al bta3 mas2la rakam 3 fe as2elet el wave fe sheet doc 3esmat
7atta ana baset 3ala el 7al elly fe malazem bashmohandes el banna bs brdo mafhmtsh ya reet 7ad yewd7ly

[جناب الكومندا المهم]

problem 1 ....diffraction in sheet??????????

[محمد حامد]

في الـ lissajous figures

ساعات بيجيب الـ radius وساعات لأ
السؤال الأول: امتى بيستخدم الـ radius؟؟...وبيجيبه منين اصلاً؟؟

السؤال التاني: في الاخر بيعمل check ويقول fx/fy=1/2 مثلاً.....بيجيبها منين دي؟؟

[ZeMa.FeRRaRi]

ياريت مسألة رقم 5 الDiffraction شيت د.عصمت
وشكرا مقدما...

[محمد عمر]

For an N-slit interference experiment. The condition for constructive interference is:

d \sin(\theta) = m \lambda

where d is the distance between the slits, m is the mode order and \lambda is the wavelength of light used.

In the problem, there are 3000 lines per cm, therefore d = 1/3000 cm and \theta = 30 ^0 at m = 3.

Therefore

\lambda = \frac{d \sin(\theta)}{m} = \frac{1}{3} \cdot \frac{10^{-2}}{3000} \sin (30^0) = 5556 \AA

For the next requirement, he wants \theta at which m=1 for the same wavelength. By substitution we find that
\sin (\theta) = \frac{m \lambda}{d} = \frac{1 \cdot 5556 \times 10^{-10}}{10^{-2} \cdot \frac{1}{3000}} \Rightarrow \theta = 9.55^0 \approx 9.6 ^0


problem 1 ....diffraction in sheet??????????

[محمد عمر]

In Lissajou's figuers, you get the radius when required in a part of the problem or required (asked for) explicitly.

It would be given in the problem statement or in terms of a constant which you would scale your answer with respect to, for example if the constant is a, then your grid should be 1a 2a 3a and so on.

for fx/fy, you know there is a relation between the number of tangents to the box surrounding Lissajous and the frequency of the signals on the x and y axes.


في الـ lissajous figures

ساعات بيجيب الـ radius وساعات لأ
السؤال الأول: امتى بيستخدم الـ radius؟؟...وبيجيبه منين اصلاً؟؟

السؤال التاني: في الاخر بيعمل check ويقول fx/fy=1/2 مثلاً.....بيجيبها منين دي؟؟

[محمد عمر]

or an N-slit interference experiment. The condition for constructive interference is:

d \sin(\theta) = m =\lambda

where d is the distance between the slits, \theta is the angle at which the maximum intensity occurs m is the mode order and \lambda is the wavelength of light used.

In the problem we have two wavelengths; \lambda_1 = 5890 \AA and \lambda_2 = 5896 \AA. For the separation between slits, we have 2500 slits in 0.5 cm, therefore d = \frac{0.5 \cdot 10^{-2}}{2500} = 2 \cdot 10^{-6} = 2 \mu m

To get the angle of the maximum of \lambda_1 and \lambda_2 we substitute

\sin(\theta_1) = \frac{1 \cdot \lambda_1}{d} = 17^09'

and

\sin(\theta_2) = \frac{1 \cdot \lambda_2}{d} = 17^08'

Therefore the angular separation \Delta \theta = \theta_1 - \theta_2 = 1'
For the resolution power. The resolution power needed for an object is given as \frac{\lambda}{\delta \lambda} while for the diffraction grating the resolution power is given as m'N.

In our problem take \lambda \approx = \lambda_1 \approx \lambda_2 or the average. \delta \lambda = \vert \lambda_1 - \lambda_2 \vert and m' = 1 and N = 2500

By substituting we will find the required RP is less than that of the diffraction grating resolving power thus they can be resolved.

ياريت مسألة رقم 5 الDiffraction شيت د.عصمت
وشكرا مقدما...

[ZeMa.FeRRaRi]

or an N-slit interference experiment. The condition for constructive interference is:

d \sin(\theta) = m =\lambda

where d is the distance between the slits, \theta is the angle at which the maximum intensity occurs m is the mode order and \lambda is the wavelength of light used.

In the problem we have two wavelengths; \lambda_1 = 5890 \AA and \lambda_2 = 5896 \AA. For the separation between slits, we have 2500 slits in 0.5 cm, therefore d = \frac{0.5 \cdot 10^{-2}}{2500} = 2 \cdot 10^{-6} = 2 \mu m

To get the angle of the maximum of \lambda_1 and \lambda_2 we substitute

\sin(\theta_1) = \frac{1 \cdot \lambda_1}{d} = 17^09'

and

\sin(\theta_2) = \frac{1 \cdot \lambda_2}{d} = 17^08'

Therefore the angular separation \Delta \theta = \theta_1 - \theta_2 = 1'
For the resolution power. The resolution power needed for an object is given as \frac{\lambda}{\delta \lambda} while for the diffraction grating the resolution power is given as m'N.

In our problem take \lambda \approx = \lambda_1 \approx \lambda_2 or the average. \delta \lambda = \vert \lambda_1 - \lambda_2 \vert and m' = 1 and N = 2500

By substituting we will find the required RP is less than that of the diffraction grating resolving power thus they can be resolved.

Thanks Alot Eng. Mohamed for your efforts
=))
god bless u