Question

Just solved (Task 2)

Just solved (Task 2)

You work as a technical engineer in a company that designs and installs sound systeins, such as speakers for large halls and open spaces. The company manager has assigned you to design a new sound system for a stadiun, and asked you to make a technical report explaining some preliminary studies on the behaviour of sound waves. Task 1: A When we are designing the sound levels to ensure a clear reception by a human ear, we often express the sound intensity in a unit called dB. The intensity level in dB of a sound wave can be found by converting the sound intensity produced by the speakers, I, expressed in Watts/m”, using the formula: B(dB) = 1010g10 Where: B is the sound intensity level in dB. I is the sound wave intensity in W/m². Io is a constant representing the intensity threshold of hearing, which is the lowest intensity a human can hear with his/her ears. A sound source, or a speaker, produces a wave with intensity I, which varies with respect to time as given below: I = 3+2 cos(4t – T)W/m². Knowing that lo = 1 x 10-12 W/m²: 1. Find a formula for the rate of change of B with respect to time. [Hint: use the chain rule] 2. Find the value of the rate of change of B with respect to time at t = 1 sec. B. The speed [v] of the sound wave travelling across the stadium is found to have the following equation: X- + 2x Where: 0<x<3.5 x is the distance travelled across the stadium in 100 meter units [x=1, means a distance of 100 meters, x =2 a distance of 200 meters, and so on), evaluate: 1. The stationary points of the velocity function for the range 0<x<3.5. 2. Define which stationary point represents a maximum velocity, a minimum velocity, and an inflexion point in the velocity within the specified period. Use higher order derivatives test for finding these points and show your solution step-by-step.

Just solved (Task 2)

Answer 1

TA w=ffde = f(covox+ sox3) de 5002? + 12.521+4 at a=0.tm at x=0, W=0 W = Sob(0,01) +1215X10 4 » G=0 = 5+ 0.00125 = 5.00025 Joule V Vang Seal cos(wtto) at - tv sin cotto = V (sin (WTED) – sind TW ino Το 2π/ω. 2 TL [si sin (21270) -sino ar [sino-sino -sino] Ivang=o -O

with Hence conside the variation. te viscosity of a lubricant decreases increase in Temperature idering I (06)] HaTTBT в Aeri я ези (1) х 10 4 аяга u = lo → ► 0,912 (1+10x+100p) = 0,384 (L+200+40mg) 0,1912 +7,126 + 7112B= 01381+ 4620 +1 x + 152,448 - 0.331 - 0.50 +8.2B. 15% + Blog 2 +162.41 = 0.662_7 -> (2) Applying equ (2) for 30 24°C = '02204 (1+300 +9008) = 0.109 (+404+1600p). → 0,204 +6,128 +18 18368 = 0.109+ 4.366+ 174 1944 13 -0,095 {at 5.220 122908 0,053972)+(3) 1,4608 +92R freno (21803) 157,1728 3=0,0043 0, 675977 =-0.036456 TN +-0.0369587 0,004372 e at 80°c 01922 (1-0,36458 +0143) 2 1- (0.036458X80)+(0.0043 X6400) u= 0.029626 D page