Question
polnts s An My Notes AskYour T A whale uses sonar (sound waves) to supplement Its eyesight. The whale swims in water where the speed of sound is measured to be 1519 m/s, and stays at rest with respect to the water Sensory Input: The whale can determine the location and speed of another object in the water, by sending out a sound wave towards it. a.) The sound returns in 0.7 seconds. How far away is the object it bounced off? 531.65 b.) The wave arrives back with its frequency Doppler shifted upwards by 2%. How fast is this object moving and in what direction? 30.38 m/s towardsve the whale Assuming the object continues with its current velocity, how much time until it arrives at the whale? 17.5 The environment: c.) Dissolving salt into water will: increase its density, which (considered alone) would tend to decrease the speed of sound (compared to in fresh water.) increase its bulk modulus, which (considered alone) would tend to decreaseX the speed of sound (compared to in fresh water.) So far, can you tell conceptually if the speed of sound is faster in salt or fresh water? d.) Assuming this salt water has a bulk modulus that is 9% higher than fresh water, find how much salt is dissolved in every cubic meter of this particular sea water aNote: Salinity varies in different parts of the world's oceans.
Explanation / Answer
a)
2 x Distance = velocity *time = 1519*0.7
So,
Distance = (1519*0.7)/2 = 531.65m
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b)
The object perceives a frequency N’ and
N’ / N = (C + v) / C
The whale perceives a frequency N” and
N”/ N’ = C /(C- v)
On multiplying the two
N”/N = (C + v) / (C-v) = 1.02 (given 2% increase)
(1519 +v) / (1519 –v) = 1.2
(1519 +v) = (1519 -v)1.2
(1519 +v) = 1822.8 – 1.2v
(1 + 1.2)v = 1822.8 – 1519
v = 138.09 m/s
_______________
c)
decrease
increase
d)
not sure
