Question #5ca90

1 Answer
ali ergin
Apr 26, 2017

#"the receiver frequency is 9907.68 Hz."#

Explanation:

#v_("air"):"velocity of sound in the air ,"v_("air")=340 " "m/s#

#v_s:"velocity of source ,"v_s=140 (km)/(hr)=(140.000)/(3600)=38.89" "m/s#

#v_r:"velocity of receiver ,"v_r=150" "(km)/(hr)=(150.000)/(3600)=41.67" "m/s#

#f_s:"frequency of source ,"f_s=10kHz=10000Hz#

#f_r:"frequency of receiver "#

#f_r=f_s * (v_("air")+-v_r)/(v_("air")+-v_s )#

#"Since the distance between receiver and source increase,"#
#"you should write the sign of " v_r " as negative and the sign of the " v_s " as positive."#

#f_r=f_s *(v_("air")-v_r)/(v_("air")+v_s)#

#f_r=10000 *((340-41.67))/((340+38.89))#

#f_r=9907.68 " "Hz#

Related questions
See all questions in Doppler Effect and Shock Wave
Impact of this question
1968 views around the world
You can reuse this answer
Creative Commons License