","comment":{"@type":"Comment","text":" Use the Doppler effect equation for the source and observer moving towards each other. Here the source is an ambulance siren and the observer is a car driver. The fundamental frequency is the frequency of the ambulance siren. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" 250 Hz","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" 500 Hz ","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" 350 Hz","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" 394 Hz","position":1,"answerExplanation":{"@type":"Comment","text":" $$f'=\\left( \\dfrac{v+{{v}_{o}}}{v-{{v}_{s}}} \\right)\\times f \\\\ \\Rightarrow f'=\\left( \\dfrac{360+50}{360-100} \\right)\\times 250 \\\\ \\Rightarrow f'=\\dfrac{410}{260}\\times 250 \\\\ \\therefore f'\\approx 394\\,Hz$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Doppler effect Quiz 1","text":" Because of jetlag, aeroplane causes a sound of frequency $$50\\,Hz$$. In airport, the crew member moving towards the plane at a speed of $$20\\,\\,{m}/{s}\\;$$ hears the jetlag sound of frequency __ $$(v=330\\,\\,{m}/{s}\\;)$$ ","comment":{"@type":"Comment","text":" Use the Doppler effect equation for an observer moving towards a stationary source, as, the source, that is, the aeroplane is stationary and the observer, that is, a crew member is moving towards the plane. The fundamental frequency is the frequency of jet lag sound. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" 40 Hz","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" 51 Hz","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" 52 Hz ","position":2}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" 53 Hz","position":3,"answerExplanation":{"@type":"Comment","text":" $$f{'}=\\dfrac{v+{{v}_{o}}}{v}\\times f \\\\ \\Rightarrow f{'}=\\dfrac{330+20}{330}\\times 50 \\\\ \\therefore f'=53\\,Hz$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Doppler effect Quiz 1","text":" A delivery boy while blowing a horn with a frequency of 150 Hz hits the wall of the house. After the accident, the frequency of horn heard by a person inside the house is 170 Hz. The speed of delivery boy before hitting the wall was $$(\\text{The speed of sound in air, }v=360\\,\\,{m}/{s}\\;)$$ $$ \\\\ $$ ","comment":{"@type":"Comment","text":" Use the Doppler effect equation for the source moving towards a stationary observer at rest, as, the source, that is, a delivery boy is moving towards the house and the observer, that is, the house is stationary. The fundamental frequency is the frequency of the horn."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":"$$\\,42.95\\,{m}/{s}$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":"$$\\,42.75\\,{m}/{s}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":"$$\\,42.55\\,{m}/{s}$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":"$$\\,42.35\\,{m}/{s}$$","position":2,"answerExplanation":{"@type":"Comment","text":" $$f'=\\left( \\dfrac{v}{v-{{v}_{s}}} \\right)\\times f \\\\ \\Rightarrow 170=\\left( \\dfrac{360}{360-v} \\right)\\times 150 \\\\ \\Rightarrow 360-v=\\dfrac{360\\times 150}{170} \\\\ \\Rightarrow v=360-317.6 \\\\ \\therefore v=42.35\\,{m}/{s}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Doppler effect Quiz 1","text":" The children running with a speed of $$20\\,\\,{m}/{s}\\;$$behind a ice cream selling vehicle moving with a speed of $$50\\,\\,{m}/{s}\\;$$. The ice cream vehicle bells with a frequency of f Hz. The children hear the frequency of the bell to be 80 Hz. Find the value of âfâ. $$(\\text{The speed of sound in air, }v=360\\,\\,{m}/{s}\\;)$$","comment":{"@type":"Comment","text":" Use the Doppler effect equation for the observer moving towards the source and the source moving away from the observer. Here the source is an ice cream vehicle and the observer is children. The fundamental frequency is the frequency of the bell. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" 66 Hz","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" 76 Hz","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" 56 Hz ","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" 86 Hz ","position":0,"answerExplanation":{"@type":"Comment","text":" $$f'=\\dfrac{v+{{v}_{o}}}{v+{{v}_{s}}}f \\\\ \\Rightarrow 80=\\dfrac{360+20}{360+50}\\times f \\\\ \\Rightarrow 80=\\dfrac{380}{410}\\times f \\\\ \\Rightarrow f=\\dfrac{80\\times 410}{380} \\\\ \\therefore f=86\\,Hz$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}