speed of sonar signal: v = 5470 km/h speed of detector: v D = 70 km/h speed of s
ID: 1724433 • Letter: S
Question
speed of sonar signal: v = 5470 km/h speed of detector: vD = 70 km/h speed of source: vs = 50 km/h Frequency of the sonar signal,f = 1000 Hz a)Frequency of the signal as detected by the U.s. subis, f ' = (v + vD) f / (v - vs) = (1000 Hz)(5470 km/h + 70 km/h) / (5470km/h - 50km/h) = 1022 Hz = 1.022 kHz b)If the French sub were stationary, thefrequency of the reflected wave would be, fr =f(v+vD)/(v-vD) But since the French sub is movingtowards the reflected signal with speed vs isthen, fr' = fr(v+vs)/v =f[(v+vs)(v+vD)/v(v-vD)] = 1.04x103HzExplanation / Answer
Given : . speedof sonar signal: v = 5470 km/h speed of detector: vD = 70 km/h speed of source: vs = 50 km/h Frequency of the sonar signal,f = 1000 Hz . (a) Form Doppler's effect : . Frequency of the signal detected is : . f ' = [ (v + vD) / (v - vs) ] *f . = [ (5470 km/h + 70 km/h) / (5470 km/h - 50km/h) ] *(1000 Hz) . = 1.0221 * (1000 Hz) . = 1.0221 kHz . (b) The French sub ismoving towards the reflected signal with speed vs isthen, . fr' = fr ( v +vs) / v --------------- (1) . But the frequency of the reflected wave would be, . fr = f (v + vD) / ( v -vD) . Substituting this value in equation (1) , we get . fr' = { [ f (v +vD) / ( v - vD) ] * ( v +vs) } / v . = [ f (v + vD) * ( v + vs) ] / [ ( v - vD)* v ] . = [ 1000 * (5470 + 70 ) * ( 5470 + 50 ) ] / [ (5470 - 70) * 5470 ] . = 1,035.3036 = 1.035 * 103 = 1.04 * 103 Hz . Hope this helps u! . Substituting this value in equation (1) , we get . fr' = { [ f (v +vD) / ( v - vD) ] * ( v +vs) } / v . = [ f (v + vD) * ( v + vs) ] / [ ( v - vD)* v ] . = [ 1000 * (5470 + 70 ) * ( 5470 + 50 ) ] / [ (5470 - 70) * 5470 ] . = 1,035.3036 = 1.035 * 103 = 1.04 * 103 Hz . Hope this helps u!Related Questions
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