Unit 4 Problems Solutions

CC2404 Applied Physics and Instrumentation in Health Care Problems Unit 4 1. i) What is the physical mechanism that causes the frequency response by the ear? ii) Outline briefly how the ear is able to distinguish sounds of different frequencies. iii) State the range of frequencies detectable by the ear for a person with normal hearing. iv) Why there is an upper limit and lower limit of the frequencies that can be detected. i) ) resonance. ii) Short stiff hairs in the cochlea resonate at high frequencies, whereas the longer fibers respond at low frequencies. The middle ear acts as a resonant cavity, giving resonances between about 700- 1500 Hz. Resonance also occurs in the ear canal (auditory canal) which acts as a tube of length about 2.5 cm, closed at one end. This gives a resonant frequency at 3 kHz. iii) 20 Hz to 20 kHz cavity to resonate.

vi) Below or above this range there is no stimulation of the hairs

2. Give one example of application of ultrasound each in therapeutic and diagnostic. What is their most important difference in the ultrasound. Therapeutic- shatter gallstones ultrasound diathermy Diagnostic- thickness measurement, bone density measurement, ultrasound imaging, their most important difference is in their intensity. 3 The threshold of hearing at a frequency of 3 kHz for a person with defective hearing is 3.2× 10-10 Wm-2. Calculate the loss, in dB, of the person's hearing. What is the sound level at 4kHz of a sound with loudness of 120 phon ?

From 0 phon curve, the sound level at a frequency of 3 kHz for a healthy hearing person is – 8dB. The sound level (intensity level) at a frequency of 3 kHz for a person with defective hearing threshold of 3.2× 10-10 W/m2, is 10 log

I 3.2 ×10 −10 = 10 log = 25 dB I0 1.0 ×10 −12

The loss is therefore: 25 dB −( −8 dB ) = 33 dB 120 phon: 110dB at 4kHz

4. a. A loudspeaker may be considered to be a point source of sound emitting sound power uniformly in all directions. The sound power produced is 0.90W. Calculate, for a point 1.5 m from the loudspeaker, i) the intensity of sound, ii) the intensity level. The surface area A of a sphere of radius r is given by A = 4π r2. i) The sound power is emitted uniformly over a sphere of radius 1.5 m. Surface area of sphere = 4π × 1.52 = 28.3 m2 power 0.90 = = 0.032 Wm −2 area 28 .3 I 3.2 ×10 −2 = 10 log = 105 dB ii) Intensity level = 10 log I0 1.0 ×10 −12 intensity =

b. Comment on your answer to a ii) with reference to any possible health hazard. Damage from several hour exposure. Long term exposure results in hearing loss.