Question #a6987

when the speeds of source and the receiver relative to the air are lower than the velocity of sound in the air, the relationship between observed frequency #nu# and frequency of sound emitting from source #nu_0# is given by Doppler's effect as follows

#nu= (v+v_r)/(v+v_s)nu_0....[1]#,

where

• #v# is the velocity of sound in the air;

• #v_r# is the velocity of the receiver of sound relative to the air. It is positive if the receiver is moving towards the source and negative in the other direction;

• #v_s# is the velocity of the source relative to the air. It is positive if the source is moving away from the receiver and negative in the reverse direction.

• Let at an instant the line joining the position of 1st car (#A#) and position of the 2nd car (#B#) makes an angle #theta# with the direction of motion of the first car. So this line will make angle #(90-theta)^@# with the direction of motion of the 2nd car.

• So the component of velocity of the 1st car along #AB# will be #v_(AB)=v_1costheta#,where #v_1# is the velocity of the 1st car towards the crossing.

• And the component of velocity of the 2nd car along #BA# will be #v_(BA)=vcos(90-theta)=v_2sintheta#, where #v_2# is the velocity of the 2nd car towards the crossing.

So

#v_r=v_2sintheta#

and #v_s=-v_1costheta#

Now the equation [1] becomes

#nu= ((v+v_2sintheta)/(v-v_1costheta))nu_0#

Now

• #v_1=72km"/"hr=20m"/"s#

• #v_2=36km"/"hr=10m"/"s#

• #v==340m"/"s#

• #nu_0=280Hz#

So

#nu= ((340+10sintheta)/(340-20costheta))280# Hz

When #theta =45^@# we finally get the frequency of horn heard by the driver of 2nd car.

#nu= ((340+10sin45)/(340-20cos45))280~~298# Hz