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Gnoma [55]
2 years ago
10

A train is traveling west at a velocity of 25 m/s. Another train is traveling east directly toward the west-bound train at a vel

ocity of 15 m/s. The west-bound train blows its whistle with a frequency of 600 Hz when the two trains are 1000 m apart and then blows its whistle again 10 seconds later. For passengers on the east-bound train, how will the perceived frequency of the first whistle compare with the perceived frequency of the second whistle
Physics
1 answer:
dusya [7]2 years ago
4 0

(a) The frequency observed by the passenger on the east-bound train when they are 1000 m apart is 675.5 Hz.

(b) After 10 seconds, the frequency observed by the passenger on the east-bound train will be greater than the frequency observed when they are 1000 m apart.

<h3>Observed frequency</h3>

The frequency observed by the passenger on east-bound train is calculated by applying Doppler shift formula.

f = f_0(\frac{v \pm v_0}{v\pm v_s} )

where;

  • f is observed frequency
  • f₀ is original frequency or source frequency
  • v is speed of sound
  • v₀ is speed of observer (positive when moving towards source)
  • vs is source frequency (negative when moving towards observer)

<h3>when the two trains are 1000 m apart</h3>

f = 600(\frac{343 +15}{343-25} )\\\\f = 675.5 \ Hz

The frequency observed by the passenger increases with decrease in distance between the two trains.

Thus, after 10 seconds, the frequency observed by the passenger on the east-bound train will be greater than the frequency observed when they are 1000 m apart.

Learn more about Doppler shift here: brainly.com/question/23841569

#SPJ1

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Answer:

c=71.4m/s

\theta=8.54\textdegree

Explanation:

From the question we are told that

Initial velocity of 60 m/s

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Generally Resolving vector mathematically

  sin(45\textdegree)15=10.6\\cos(45\textdegree)15=10.6

Generally the equation Pythagoras theorem is given mathematically by

c^2=a^2+b^2

c^2=10.6^2 +(10.6+60)^2

c=\sqrt{10.6^2 +(10.6+60)^2}

Therefore Resultant velocity (m/s)

c=71.4m/s

b)Resultant direction

Generally the equation for solving Resultant direction

\theta=tan^-1(\frac{y}{x})

Therefore

\theta=tan^-1(\frac{10.6}{70.6})

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A 0.50-kg mass attached to the end of a string swings in a vertical circle (radius 2.0 m). When the mass is at the highest point
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You are traveling in a car toward a hill at a speed of 36.4 mph. The car's horn emits sound waves of frequency 231 Hz, which mov
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Answer:

<em>a. The frequency with which the waves strike the hill is 242.61 Hz</em>

<em>b. The frequency of the reflected sound wave is 254.23 Hz</em>

<em>c. The beat frequency produced by the direct and reflected sound is  </em>

<em>    11.62 Hz</em>

Explanation:

Part A

The car is the source of our sound, and the frequency of the sound wave it emits is given as 231 Hz. The speed of sound given can be used to determine the other frequencies, as expressed below;

f_{1} = f[\frac{v_{s} }{v_{s} -v} ] ..............................1

where f_{1} is the frequency of the wave as it strikes the hill;

f is the frequency of the produced by the horn of the car = 231 Hz;

v_{s} is the speed of sound = 340 m/s;

v is the speed of the car = 36.4 mph

Converting the speed of the car from mph to m/s we have ;

hint (1 mile = 1609 m, 1 hr = 3600 secs)

v = 36.4 mph *\frac{1609 m}{1 mile} *\frac{1 hr}{3600 secs}

v = 16.27 m/s

Substituting into equation 1 we have

f_{1} =  231 Hz (\frac{340 m/s}{340 m/s - 16.27 m/s})

f_{1}  = 242.61 Hz.

Therefore, the frequency which the wave strikes the hill is 242.61 Hz.

Part B

At this point, the hill is the stationary point while the driver is the observer moving towards the hill that is stationary. The frequency of the sound waves reflecting the driver can be obtained using equation 2;

f_{2} = f_{1} [\frac{v_{s}+v }{v_{s} } ]

where f_{2} is the frequency of the reflected sound;

f_{1}  is the frequency which the wave strikes the hill = 242.61 Hz;

v_{s} is the speed of sound = 340 m/s;

v is the speed of the car = 16.27 m/s.

Substituting our values into equation 1 we have;

f_{2} = 242.61 Hz [\frac{340 m/s+16.27 m/s }{340 m/s } ]

f_{2}  = 254.23 Hz.

Therefore, the frequency of the reflected sound is 254.23 Hz.

Part C

The beat frequency is the change in frequency between the frequency of the direct sound  and the reflected sound. This can be obtained as follows;

Δf = f_{2} -  f_{1}  

The parameters as specified in Part A and B;

Δf = 254.23 Hz - 242.61 Hz

Δf  = 11.62 Hz

Therefore the beat frequency produced by the direct and reflected sound is 11.62 Hz

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