9. There are 3 trumpet players, each playing with sound intensity level of 90 4B

Question

9. There are 3 trumpet players, each playing with sound intensity level of 90 4B. What will be the total sound intensity level when all 3 are playing together? Ignore any immerference 10. An instrument is being played at a frequency of 200 Hz. a. What is the frequency of a note that is an octave higher?H2 H b. an octave lower? les Ha scale, the interval is 1.05946. Starting with a frequency of 600 Hz a, what is the frequency of the next higher note? ooort b. what is the frequency of the next lower note to 500 He? u 12 Suppose you cover up the bell of a woodwind instrument (such as a clarinet) and of each instrument. What kind of instrument (eg, a trumpet) and blow through the mouthpiece sound will be produced from each instrument? Briefly explains your answer 13. Consider 2 waves that are composed of same number of harmonics of same amplitudes. The only difference between the 2 waves is the phase of their harmonics. Will your ear notice any difference in sound heard from the 2 waves? Also, state the scientific lavw/principle your answer based on. frequency and 14. Consider the following pairs of frequencies. i. 200 Hz, 304 Hz ii. 303 Hz, 300 Hz iii. 400 Hz, 800 Hz a. Which pair will produce a beat frequency? 303 Hz ,oo Hz b. What is the beat frequency? 1- 7H-onHh2H c. Can you provide any theory as to why a beat frequency sounds unpleasant? 15. Sound produced by the first harmonic (the fundamental frequency) of a vibrating string sine wave that is shown below, Show the Fourier transform of this wave on the plot on the hand side. 2

Explanation / Answer

Sound intensity level is measured in dB, while intensities are not

for a source with intensity I1, the level is defined as L1 = 10log(I1/I0)

when three sources are emitting sounds together, the intensity values are added up

hence, I = I1 +I2 +I3

L1 = 90dB

90 = 10log(I1/I0)

I1 =Io *109

when three trumpets are playing together, intenisty = 3I1

I =3Io *109

sound level = L = 10log (3Io *109 /I0)

L = 10log3 +10logI1

=4.77 + 90

L = 94.77dB --------answer

2. a) an octave higher is the twise the frequ3ency, hence, the new frequency will be 400Hz

b) octave lower is the half of original frequency, hence new frequency will be 100 Hz

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