Industrial Noise and Vibration

صفحه 1:
Industrial Noise and a Vibration Solution for Session 2 Physical Acoustics mmonazzam@hotmail.com 

صفحه 2:
Answer (example (1)) = If the displacement of the particles of the medium is described by: d=0005i4 20rt- x+ a * What is the amplitude, frequency wavelength and wave number and what is the speed of the wave? = Solution: = From the above equation the amplitude is 0.005 wo =2nf =20r ‏م‎ ۶-20 =10 He بر 7 ‎=20r‏ 21027 دجم 

صفحه 3:
Answer (example (2)) = The pressure fluctuations in air are described by: p=0.01co$20Grt- 1.85x) + 0.0055ir200rt- 1.859) = What is the amplitude, frequency, wavelength and the speed of the wave. = Solution: |g =V00F + 0.005 = | =0.011 pa ‏هه‎ =2nf =200r = £=200/ =100 Hz 27 00 we Ka> =185 A= 7 ‏سوع‎ A =34 metre. = A= c=100kK34 =340m7/s 

صفحه 4:
Answer (example (3)) © The pressure, p, is described by: 2 ‏مم در‎ ۵ If the pressure amplitude is 0.01 pa and at t=0 , x=0 the value of p is 0.005 pa find Ar and A\. Solution: The expression for the pressure is: p=Deokut- ko) D=|#+# — and ta =4 thereforatt=0&x=0 p=0005-Deow, and we know D=001 005-95 9 =60 therefore cos) =: tar =1.73 A+ # 17 A =A tar =0.0054.73=0.00865 pa = A(1+1.73)=0.0001 and 4 =0.005 pa 

Industrial Noise and Vibration Solution for Session 2 Physical Acoustics mmonazzam@hotmail.com Answer (example (1))  If the displacement of the particles of the medium is described by:   d 0.005sin 20t  x    2  What is the amplitude, frequency ,wavelength and wave number and what is the speed of the wave?  Solution: From the above equation the amplitude is 0.005 m  2f 20  f 20 2 10 Hz 2 k 1  2 and   c  f  c 102 20 Answer (example (2))  The pressure fluctuations in air are described by: p 0.01cos200t  1.85x  0.005sin200t  1.85x  What is the amplitude, frequency, wavelength and the speed of the wave.  Solution: p  0.012  0.0052  p 0.011 pa  2f 200  f 200 2 100 Hz k 2 1.85  2   3.4 metres 1.85  c  f  c 1003.4 340m/ s Answer (example (3))  The pressure, p, is described by: j t kx  p  Ae  If the pressure amplitude is 0.01 pa and at t=0 , x=0 the value of p is 0.005 pa find Ar and Ai. Solution: The expression for the pressure is: p D cost  kx   D  Ar2  Ai2 and tan  Ai Ar therefore at t 0&x 0 p 0.005D cos , and we know D 0.01 therefore cos  0.005 0.5   600 0.01  tan 1.73 2 2   A2  A2 0.012  Ar 1 1.73 0.0001 and Ar 0.005 pa i  r Ai  Ar tan 0.0051.730.00865 pa  