Answer:
Therefore,
The frequency heard by the engineer on train 1
[tex]f_{o}=603\ Hz[/tex]
Explanation:
Given:
Two trains on separate tracks move toward each other
For Train 1 Velocity of the observer,
[tex]v_{o}=145\ km/h=145\times \dfrac{1000}{3600}=40.28\ m/s[/tex]
For Train 2 Velocity of the Source,
[tex]v_{s}=90\ km/h=90\times \dfrac{1000}{3600}=25\ m/s[/tex]
Frequency of Source,
[tex]f_{s}=500\ Hz[/tex]
To Find:
Frequency of Observer,
[tex]f_{o}=?[/tex] (frequency heard by the engineer on train 1)
Solution:
Here we can use the Doppler effect equation to calculate both the velocity of the source [tex]v_{s}[/tex] and observer [tex]v_{o}[/tex], the original frequency of the sound waves [tex]f_{s}[/tex] and the observed frequency of the sound waves [tex]f_{o}[/tex],
The Equation is
[tex]f_{o}=f_{s}(\dfrac{v+v_{o}}{v -v_{s}})[/tex]
Where,
v = velocity of sound in air = 343 m/s
Substituting the values we get
[tex]f_{o}=500(\dfrac{343+40.28}{343 -25})=500\times 1.205=602.64\approx 603\ Hz[/tex]
Therefore,
The frequency heard by the engineer on train 1
[tex]f_{o}=603\ Hz[/tex]
