Chapter 10 Question 1
 
a. With reference to the sound produced by a loudspeaker, explain why the frequency of the emitted sound waves is the same as the frequency of the vibrating source. 3 marks
 
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When the diaphragm of the loudspeaker moves to the right, a compression is produced. When it moves to the left, a rarefaction is produced. 1
A complete sound wave (consisting of a compression and a rarefaction) is generated when the diaphragm moves to-and-fro in one complete cycle. Thus, the frequency of sound is equal to the frequency of the diaphragm. 1
 
b. What is an “ultrasound”? Give a brief account on the medical diagnosis using ultrasonic imaging. Compare this with X-ray imaging. 4 marks 
 
Ultrasounds are sound waves that have a frequency higher than 20 kHz. 1
In ultrasonic imaging, the ultrasound transducer is placed on the surface of the patient, with a thin layer of liquid in between. 0.5
Pulses of ultrasound are emitted and penetrate into the body. They are reflected from different depths within the body. A computer converts the reflected signals into a picture that displays the internal organs for diagnosis. 0.5
In X-ray diagnosis, the detector is located on the opposite side as the transmitter to detect the transmitted signal; while in ultrasound imaging, the reflected signal is detected. 0.5
X-rays are high energy radiations and are harmful to human body. Ultrasounds are harmless. 0.5
The penetration power of X-rays is much higher than ultrasound. They are absorbed by high density material. Thus, X-rays are good for revealing bone fractures but cannot be used to detect tissue organs. 1
 
c. With the help of a diagram, give a brief account on how ultrasound can be used to detect flaws in materials. 4 marks
 
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A sonar transmitter and detector is placed in contact with one end of the material. 0.5
A sharp pulse of ultrasound is sent toward the material. 0.5
If the material is perfect, the video screen will show two pulses, one due to the transmitted pulse, the other due to the reflection at the far end. 1
If there is a flaw in the material, the flaw will reflect partially. Thus, the video screen will give more than two pulse as shown. 1
 
d. Explain the meaning of “loudness”. Give three physiological factors that may affect the perception of intensity of sound. 5 marks
 
Loudness is the physiological perception of sound intensity. It is a sensation of the human ear, varying from one person to another. 1
Our ear responds logarithmically to the intensity. If the sound intensity level increases by 20 dB, the new sound seems approximately twice as loud as the original sound. 1
An increase in 20 dB is in fact an increase of intensity by hundred (100 = 102) times. 0.5
0.5
Our hearing system responds differently to different frequencies. We are more sensitive to sound with frequencies 1 kHz to 5 kHz. 1
Another factors is the duration. Long bursts of sound are louder than the short bursts. 1
 
 
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