Question

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 velocity 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

Answer 1

(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.

Observed frequency

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)

when the two trains are 1000 m apart

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

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: https://brainly.com/question/23841569

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