物理計算題 ~20點

2015-01-19 2:33 am
Q1 Cave swiftlets are a type of small cave dwelling birds. Like bats, cave swiftlets use echo location to find their way in the the dark. To warn it's family of its approach a certain swiftlet sends out a sound pulse with a frequency of 9.05 kHz. The swiftlet if flying towards the nest with a speed of 12.1 m/s. The speed of sound in air is 343 m/s.

1a)What frequency pulse does the family in the nest detect?f' = __kHz

1b)What frequency pulse does the swiftlet detect reflected from the nest?f" =__kHz

1c)On detecting the pulse the parent waiting in the nest realises that it is its turn to leave to find food. It flies towards the approaching swiftlet with a speed of 12.7 m/s. What frequency pulse does the swiftlet leaving the nest now detect coming from the swiftlet approaching the nest? f' =__ kHz

1d)What frequency pulse does the swiftlet approaching the nest detect reflected of the swiftlet leaving the nest? f" = __kHz

Q2 The body of a car is suspended on four identical springs. Consider a car with a mass of 1475 kg. This mass is evenly distributed over the four springs. These springs cause the car to oscillate with a frequency of 2.51 Hz when empty.

2a) What is the spring constant of these springs?__ N/m
2b)Five large people, each with a mass of 111 kg climb into the car. What distance does the suspension of the car drop? __cm
2c) What frequency does the car oscillate with now? __Hz
更新1:

~~~Please help~~~thank you~~~

更新2:

謝謝~~ 不過我交了first draft 有兩題答案不對 1d)What frequency pulse does the swiftlet approaching the nest detect reflected of the swiftlet leaving the nest? f" = __kHz 2b)Five large people, each with a mass of 111 kg climb into the car. What distance does the suspension of the car drop? __cm 會有別的想法嗎?

回答 (1)

2015-01-20 12:14 am
✔ 最佳答案
1.(a) Apparent frequency = 9.05 x (343/(343 - 12.1) kHz = 9.38 kHz

(b) Reflected frequency = 9.38 x (343 + 12.1)/343 kHz = 9.71 kHz

(c) Apparent frequency = 9.05 x (343 + 12.1)/(343 - 12.7) kHz = 9.73 kHz

(d) Reflected frequency = 9.73 x (343+12.1)/ 343 kHz = 10.07 kHz

2(a) Angular frequency = 2.pi x 2.51 s^-1 = 15.77 s^-1
Spring constant k = 12.77^2 x 1475/4 N/m = 91714 N/m

(b) Total mass of car and men = (111 x 5 + 1475) kg = 2030 kg
Mass acted on a spring =2030/4 kg = 507.5 kg
Compression of each spring = 507.5g/91714 m = 0.0553 m = 5.53 cm
[Take g, the acceleration due to gravity, to be 10 m/s^2]

(c) New frequency = 1/(2.pi) x square-root[91714/507.5] Hz = 2.14 Hz

2015-01-20 16:23:47 補充:
1(d) f" = 9.73 x (343 + 12.1)/(343 - 12.7) kHz = 10.46 kHz

2015-01-20 16:44:31 補充:
2(b) Original compression of spring = 1475g/(4 x 91714) m = 0.0402 m
Final compression of spring = 507.5g/91714 m = 0.0553 m
Hence, suspension drops = (0.0553 - 0.0402) m = 0.0151 m = 1.51 cm


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