In this problem on the topic off waves and sound we're told that a car is parked given distance directly south of a railroad crossing. There is a train approaching the crossing directly from the west, headed directly east at a given speed. He trained them sounds its horn which has a given frequency when it reaches 20 m west of the crossing. We want to know the frequency that the car's driver cars drive over here when the horn blast reaches the car where given the speed of sound. And we have to assume that the only component of the trains velocity that is directed towards the car will affect the frequency heard by the driver. Now if the train were they headed directly towards the car, the frequency observed Oh would be given by Fo is equal to F s over one minus V Train of V where s the frequency of the source? The speed of sound and the train the speed of the train. However, the driver is 20 m south of the crossing and the train is 20 m where 20 m west of the crossing now the train would have to be here directly southeast in order to be moving directly towards the cards. Instant. So to calculate the Doppler shift in frequency of the horn, we have to only consider the southeast component off the trans Velocity V s E. Where the train velocity is being trained. Now, according to the drawing, we have done the scale a component of the S E off the trains, velocity is given as follows. So the component of the trains velocity to the southeast with cais park, the S E is equal to the speed of the train The chain times the call sign of 45 degrees this by trigonometry. So the observed frequency by the driver in the stationary car is equal to the frequency off the trains on F s into one over one minus the speed off the source The speed of the train towards the southeast, over the speed of sound V. So we can write this as the frequency off the sauce, which is the frequency off the train for the whole of the train s into one over one, minus the speed off the train. More times the co sign of 45 degrees. So we replace VSC Bye. the train co sign 45 degrees divided by the speed of sound V. Now all of these values are known so we can calculate the the frequency that the driver in the parked car will observe. So the frequency of the sauce is 289 huts and this is into one over one minus. The speed of the train is 55 m per second and multiplied by the coastline of 45 degrees, gives us the component towards the vodka divided by the speed of sound 343 m per second. So if we calculators, we get the frequency observed by hot car to be 326 cuts, yeah.