The motion detector we used in class measures distance to the nearest object by

Question

The motion detector we used in class measures distance to the nearest object by using a speaker and a microphone. The speaker clicks 30 times a second. The microphone detects the sound bouncing back from the nearest object in front of it. The computer calculates the time delay between the making of the sound and receiving the echo. It knows the speed of sound (about 343 m/s at room temperature) and from that, it can calculate the distance to the object from the time delay.

A) If the nearest object in front of the detector is too far away, the echo will not get back before a second click is emitted. Once that happens, the computer has no way of knowing that the echo isn't an echo from the second click and the detector doesn't give correct results anymore. How far away does the object have to be before that happens? (Give your answer to three significant figures.)

Explanation / Answer

from the given data
Time period, T = 1/30 s

= 0.03333 s

let d is the distance between the speaker and the object.

so, distance travelled by sound in one time period = 2*d

now use, 2*d = v*T

d = v*T/2

= 343*0.03333/2

= 5.72 m <<<<<<<<<<--------------------Answer

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