The Acoustic Properties of a Lecture Theatre - Case Study Example

Sound Insert Name Insert Name of University Insert Name of Course Insert Name of Professor April 10, 2014 Table of Contents Introduction 3 Room description 3 Procedure 3 Computer code 3 Results of reverberation time and discussion 3 Discussion and conclusion 11 Introduction The purpose of this study was to investigate the acoustic properties of a lecture theatre for the engineering class where audio recording is made. The room is structured in a manner that it hosts a number sessions for engineering related course it as a capacity of holding 50 sitting occupants. This room was used for this experiment in recording sound. Room description The room is hexagonal with 2 parallel walls which are good in discouraging standing waves when the experiment is being carried out. It has sides which are not equal and made of blocks that are treated with acoustic materials. It as height 7.2ft Procedure During testing the sound was made from the centre of the room mentioned above where the speaker was making various sounds and was being recorded. The student used earphones to attenuate the sound by 50% . Computer code MATLAB software used to generate codes that were used in the experiment. Results of reverberation time and discussion When SPL in a room is 140µPa, the Sound level in dB To calculate the sound level in dB, the following formulae is used Lp = 20log10 () dB Then this is applied to the SPL in a room is 140µPa Lp = 20log10 () Db = 16.901dB The SPL in the room in is doubled, 280µPa what is the increase in sound level in dB To calculate the sound level in dB, the following formulae is used Lp = 20log10 () Db Lp = 20log10 () Db Lp = 20log10 () Db = 22.923 dB 3.) SPL in the room in question 1 is increased ten-fold (1400µPa), what is the increase in sound level in dB To calculate the sound level in dB, the following formulae is used Lp = 20log10 () Db Lp = 20log10 () Db = 36.902dB 4. A manufacturer of noise-cancelling headphones claims that their product can reduce the level of background noise by 15 dB. If the level of background noise is 200 mPa, calculate the SPL when the noise reduction headphones are used. To calculate the sound level in dB, the following formulae is used Lp = 20log10 () Db Lp = 20log10 () Db 200 mPa = 200,000µPa Lp = 20log10 () Db = 80dB Reducing by 15dB is 80dB -15dB = 65dB 65dB = 20log10 () Db 3.25 = log10 () 1778.28 = () 1778.28 = () 35,565.59µPa = 35. 5. During a class when the lecturer was demonstrating various sounds, a student, trying to catch-up on sleep, uses her earphones to attenuate the “annoying” sounds made by the lecturer. She finds that this reduces the sound pressure level by 50%. What is the reduction in dB? The noise made by the lecturer is 87dB 87dB = 20log10 () dB 4.35 = log10 () 447,744.2277 µPa Reducing by 50% 50% OF 447,744.2277 = 223.872.1139 Lp = 20log10 () dB Lp = 20log10 () dB = 80.98 dB Reduction in dB is 87-80.98 dB = 6.02dB 6. The results of an experiment to measure the reverberation time of a room is shown below. a. Select the history time is 14.5 seconds suitable range to estimate the slope of the magnitude time history b. Estimate the slope of the magnitude time history =1.33dB/s c. What are the units, if any, for the slope of the magnitude time history? dD/s Db/s 7. The acoustic waveform shown in the previous question was submitted to spectral analysis resulting in the sound spectrum shown below. Identify a. The range of frequencies (apart from background noise) contained in the acoustic waveform is 1200 to 11,000. b. The type of sound longitude and transverse waves- Mechanical Wave Longitudinal Wave Pressure Wave sound wave is a longitudinal c. The centre frequency describing this type of sound is 400 8. Complete the following Matlab command: [z sr] = w _ _ r _ _ _ (‘noise.wav’); and describe the content of the variables z and sr. s = wavread (’noise.wav’); sm = s - mean(s); plot (sm); sound (sm); sr = frame (sm, 160, 0); % no overlap ! % one frame to play with x = sr (:,7); plot (x); 9. You are using Matlab and have uploaded an acoustic waveform that was recorded with a sampling frequency sr. Write the code to generate a suitable (elapsed) time vector and plot the acoustic waveform vs elapsed time. [y, fs]=wavread('noise.wav'); sound(y, fs); time=(1:length(y))/fs; plot(time, y); 10. You are using Matlab and are required to find the slope of the straight line of best fit for a magnitude vs time plot show below. However, only data between 14.5 seconds and 16.0 seconds (inclusive) must be used. Given that the time interval is 50 ms, write the code to extract the slope. 12.5 function [p1,p2,slope,inter]=getslopeintercept % SYNTAX: plot([14.5:16],[100:35].^0.5,'-o') % [p1,p2,slope,inter]=getslopeintercept % OUTPUTS: % P1= coordinates of the first cliked point % P2= coordinates of the second cliked point % SLOPE= slope of the drawn line % INTER= intercept Y(X=0) 11. Define the lower and upper frequency limits for an octave band with a centre frequency of 4,000 Hz. 20 Hz to ~ 20 KHz can be divided up into ~ 11 octave bands Lower Band Limit is 2840 and Upper Band Limit is 5680 12. The frequency of a harmonic sound of 220 Hz is increased by 2 octaves. What is the new frequency? In order to increase harmonic by 2 octaves the following formula f(N) = N fo Where fn is new frequency of harmonic sound, N is the number of octaves, F0 is the original frequency of harmonic sound. Therefore the new frequency will be calculated as follows: f(N) = 2 220 Hz = 440Hz Discussion and conclusion Sound intensity level is useful because it roughly approximates the way we perceive loudness since it is a logarithmic function of intensity. Equal increments in intensity level roughly correspond to equal increases in loudness. Two useful rules of thumb: every time the intensity increases by a factor of 10, the intensity level adds 10db; since log 10 2 0.30, adding 3.0 db to the intensity level doubles the intensity. When both lathes are running at the same time, the intensity is twice as big as for one lathe, but the two do not sound twice as loud as one. Intensity level is a better guide to loudness; two lathes produce a level 3 db higher than one lathe. Decibels can also be used in a relative sense; instead of comparing intensity to l0, we can compare two intensities directly. Suppose we have two intensities I1 and I2 and two corresponding intensity levels β1 and β2. Then Β2-β1=10dB Since log x-log y=log β2 –β1 = (10 dB) Log10 = (10dB) LOG 10 While loudness is most closely correlated to intensity level, it also depends on frequency. In other words, the sensitivity of the ear is frequency dependent. Each curve shows the intensity levels required so that sounds of different frequencies are equally loud . In this frequency range, 1 Db is about the smallest change in intensity level that is perceptible as a change in loudness. The threshold of hearing is shown by the lowest curve in the set; a person with excellent hearing cannot hear sounds with intensity levels below this curve. The threshold of hearing is at an intensity level of db only in the vicinity of 1 kHz. References Giambattista, A. ( 2009). Student Solutions Manual to accompany Physics. New York: McGraw-Hill Giambattista, A., Richardson, B. M & Richardson, R.C. (2009). College physics. New York: McGraw-Hill. Lon Turnbull, L. & David Wagoner, D. (2004). Physics 2111 Laboratory Manual Wilson, J., (1994). Physics Laboratory Experiments. New York: Houghton Mifflin Company Serway R & Jewett, J., 2008. Physics for Scientists and Engineers /With Modern Physics. New York: Thomson Read More