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ki77a [65]
3 years ago
9

A train is traveling at 30.0 m/sm/s relative to the ground in still air. The frequency of the note emitted by the train whistle

is 262 HzHz. The speed of sound in air should be taken as 344 m/sm/s.A. What frequency fapproach is heard by a passenger on a train moving at a speed of 18.0 m/s relative to the ground in a direction opposite to the first train and approaching it?B. What frequency frecede is heard by a passenger on a train moving at a speed of 18.0 m/s relative to the ground in a direction opposite to the first train and receding from it?
Physics
1 answer:
Zigmanuir [339]3 years ago
8 0

Answer

given,

speed of sound = 344 m/s

speed of train = 30 m/s

frequency emitted by the train = 262 Hz

   Doppler's effect

    f_L = \dfrac{v + v_L}{v + v_s}\ f_S

f_L is the frequency of listener

f_S is the frequency of the source of the sound

v is the speed of the sound

v_L is the speed of listener.

v_S is the speed of the source

a) Speed of the passenger in another train , v = 18 m/s

   another train is moving in opposite direction and approaching

   v_L is positive as the listener is moving forward.

    v_S is negative at the source is moving toward the listener.

      f_L = \dfrac{344 + 18}{344 - 30}\times 262

     f_L = 302\ Hz

b) Speed of the passenger in another train , v = 18 m/s

   another train is moving in opposite direction and receding

    v_L is negative as the listener is moving away from source.

    v_S is positive at the source is moving away the listener.

      f_L = \dfrac{344 - 18}{344 + 30}\times 262

     f_L = 228.37\ Hz

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During a rodeo, a clown runs 7.7 m north, turns 49.9 degrees east of north, and runs 6.4 m. Then after waiting for the bull to come near, the clown turns due east and runs 19.8 m to exit the arena. The magnitude of the clown’s displacement is 27 m.

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In first step, the clown runs 7.7 m in north direction, so the image will be  as in fig 1. Then he takes a direction of north east and covers a distance of 6.4 m, so the image will be modified as in fig 2. Then after the bull comes, he turns east and runs 19.8 m to exit the arena, so the image will be as in figure 3.

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Similarly, the adjacent side of this imaginary triangle is the distance covered by the clown in the North direction.

\text { Distance covered by the clown in north direction }=6.4 \sin 49.9+7.7=12.6 \mathrm{m}

Thus, the total displacement covered by the clown is

\text { Total displacement }=\sqrt{(23.9)^{2}+(12.6)^{2}}=\sqrt{571.21+158.76}=\sqrt{729.97}=27 \mathrm{m}

Thus, the total displacement by the clown is 27 m.

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