Question

2. A note middle C from an oboe has a frequency of 261 Hz and travels at a speed of 343 m/s. The maximum amplitude of the vibration of an air molecule as the sound goes by is 3.0 x10-5 m. One observer in a concert hall is 15 m away from the oboe while another observer at the back of the concert hall is 75 m away. (a) Determine the wavelength of the wave. (b) Determine the maximum speed of the vibration of an air molecule as it vibrates to transmit the sound. (c) Determine the difference in time that the first observer hears the oboe? f. = 261H2 Ve= 343 m/s A = 3.0x105m observer, d= 15m ob, d= 75m + 3 (2105) 2

Answer 1

a) wavelength λ = v/f

where v = wave speed = 343 m/s , f = frequency = 261 Hz

hence λ = 343/261 = 1.31 m

hence wavelength of sound = 1.31 m

b) max. Particle speed v_m = A*ω

where A = amplitude of particle = 3*10^-5 m, ω = angular frequency = 2πf

v_m = Aω = 2πAf = 2*3.14*3*10^-5*261 = 0.0492 m/s

hence max. Particle speed = 0.0492 m/s

c) distance between both observer is Δd = 75 - 15 = 60 m

speed of sound v = 343 m/s

Hence time difference in hearing sound Δt = Δd/v = 60/343 = 0.175 s

hence the time difference in hearing sound between 1st and 2nd observer is 0.175 s