Question

Two trains on separate tracks move toward each other. Train 1 has a speed of 109 km/h; train 2, a speed of 99.0 km/h. Train 2 blows its horn, emitting a frequency of 500 Hz. What is the frequency heard by the engineer on train 1?

Answer 1

The frequency heard by the engineer on Train 1 is approximately 422.95 Hz.

Step-by-step explanation:

When two objects are moving towards each other, the frequency of the sound heard by an observer on one of the objects is higher than the actual frequency of the sound. This phenomenon is known as the Doppler effect.

In this case, Train 1 and Train 2 are moving towards each other. Train 2 blows its horn at a frequency of 500 Hz. The engineer on Train 1 will hear a frequency that is higher than 500 Hz. To calculate the frequency heard by the engineer on Train 1, we can use the Doppler effect formula:

f' = f * (v + ve) / (v + vs)

Where:

• f' is the frequency heard by the engineer on Train 1
• f is the frequency of the sound emitted by Train 2 (500 Hz)
• v is the speed of sound (approximately 343 m/s)
• ve is the velocity of the engineer on Train 1 (-109 km/h)
• vs is the velocity of Train 2 (99.0 km/h)

Converting the velocities to m/s:

• ve = -30.3 m/s
• vs = 27.5 m/s
  

Plugging in the values:

• f' = 500 * (343 + (-30.3)) / (343 + 27.5)

Simplifying the equation:

• f' = 500 * 312.7 / 370.5
• f' = 422.95 Hz
  

Therefore, the frequency heard by the engineer on Train 1 is approximately 422.95 Hz.

Answer 2

`f_o=592.36 Hz`

Step-by-step explanation:

Given that

Train 1 (observer):

Speed = 109 km/h

Train 2 (source):

Speed = 99 km/h

Train 2 emitting frequency = 500 Hz

We know that observer and source are moving toward each other, then frequency heard by observer can be given as follows

`f_o=\left((C+V_o)/(C-V_s)\right)f_s`

Where C is the velocity of sound (C=1225 Km/h)

Now by putting the values

`f_o=\left((C+V_o)/(C-V_s)\right)f_s`

`f_o=\left((1225+109)/(1225-99)\right)* 500`

`f_o=592.36 Hz`