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A cannon is used to launch a speaker emitting 3 kHz waves with a vertical velocity of 80 $\mathrm{m} / \mathrm{s} .$ There is a stationary detector on the ground by the cannon.(a) What is the frequency of the waves received by the detector 10 $\mathrm{s}$after the speaker is launched?(b) If the speaker begins simultaneous emitting 3.6 $\mathrm{kHz}$ waves, what will be the value of the beat frequency produced at the speaker?(c) If the intensity of the sound received by the detector when the speaker reaches its maximum height is $8 \times 10^{-9} \mathrm{W} / \mathrm{m}^{2}$ , what is the intensity received by the detector when the speaker is at half of its maximum height?

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Chapter 13

Practice Test 3

Section 2

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Hi In the given problem, original frequency of the sound emitted by the speaker, which is thrown vertically upward is three kilo purse. No, but initially speed with which the speaker was projected vertically upward is we i is equal to 80 m/s and that is vertically upward. Now, in the first part of the problem, we have to find the apparent frequency as heard by the observer at the ground after 10 seconds. So first of all, we will have to find the speed of this speaker after 10 seconds, for which we will use first equation of motion, which says the final speed via physical to initially speed V plus G means acceleration to the gravity into time. So this we have here will come out to be for v. This is 80 as the speaker is going up. So acceleration for it will be negative since this is 10 m per second square, multiplied by 10. So it comes out to be a t minus 100 which is minus 20 m per second here. This negative sign shows that actually the speaker has started coming back vertically downward after 10 seconds means the speaker is approaching now approaching the observer with our speed and as the speaker is the source of sound, so we will symbolise this speech of the speaker as V. S. And that is 20 m per second. Speed of the observer will be taken as zero as it is addressed on the ground Speed of sound can be assumed to be 343 m/s hands using Dobler's effect, the apparent frequency as her by the observer and the ground will be given by and dash is equal to end C. Plus well divided by C minus V. S. So plugging in all non values here for N. This is three kg hertz, C is 343 plus zero, divided by 3 43 minus 20 m per second. So finally this frequency, it comes out to be 3.185 kg heart, which is the answer for the first part of the problem. Now in the second part of the problem, it is said that the speaker produces one more simultaneous frequency we can say to be and to which is 3.6 kg hertz. Why the original frequency was anyone is equal to three kg hertz. So we have to find the beat frequency as heard by the observer. So beat frequency means to change in the frequency as heard by the observer. So that change in the frequencies are heard by the observer will be equal to the change in the original frequencies. So did frequency here Will be equal to and to -N1 means this is 3.6 -3.0 kilo hertz. Or we can see This is 0.6 kilo hertz. Or finally we can see this is 600 hers. The answer for the second part of the problem. Now, in the third part of the problem, when the speaker is at the top affects vertical motion, means at the maximum height, at maximum height intensity of the sound heard by the observer. We can assume this maximum height to be edged and intensity means Ivan is eight into 10 days par minus nine what per meter square. Now we have to find the intensity heard by the observer at a height half to that of the maximum height. Hence if each one is edge, H two will be etched by two and I won is eight into 10 days par minus nine what per meter square. So edge too. So we have to find I do for which we use the concept that intensity of sound depends is universally on the square of distance means r squared. So using that we can see I won by two is equal to H two square by H one square. So for I one this is 18 to 10 days par -9. What per meter squared divided by two, which is missing is equal to edge two square means edge by two to the whole square, divided by the square, canceling the status where we will get this to be won by four. So finally I too is four times off. Ivan means 18 to 10 days par minus nine what per meter square. So finally intensity heard by the Observer renda Speaker is at a height equal to half of the maximum height comes out to be 3.2 into 10 days. To the power minus eight What per meter square, which is the answer for that? Her part of this problem. Thank you.

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