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Question 8 An observer is standing at a position 10 m from wall Between the observer and the wall, tuning fork emitting sound waves of frequency 500 Hz is moving at...

Question

Question 8 An observer is standing at a position 10 m from wall Between the observer and the wall, tuning fork emitting sound waves of frequency 500 Hz is moving ata speed Vs 2.0 m/s toward the wall How many beats per second the observer hears due to the interference between the direct and reflected sound? The speed of sound is 340 m/s.

Question 8 An observer is standing at a position 10 m from wall Between the observer and the wall, tuning fork emitting sound waves of frequency 500 Hz is moving ata speed Vs 2.0 m/s toward the wall How many beats per second the observer hears due to the interference between the direct and reflected sound? The speed of sound is 340 m/s.



Answers

A source of sound waves of frequency $1.0 \mathrm{kHz}$ is stationary. An observer is traveling at 0.50 times the speed of sound. (a) What is the observed frequency if the observer moves toward the source? (b) Repeat if the observer moves away from the source instead.

Hello students. So in this question a sound of frequency 5 30 hertz. That is FS. Equals 25 30 hertz is amateur by agitation. Every source of yeses equals to zero m per second. Okay. And observer is approaching the source so we is give given and it is approaching towards the stationary source with high speed, received the sound and measured it as if not equals 25 80 hertz. Okay, so for the part we have to determine the speed of the observer so we can write for this condition if not, will be equal to F. S. Medical herb. We plus we oh dear. Where we where we is the speed of sound which is given as 3 40 m per second. So substituting the value so we will get here if not which is 5 80 Fs which is 5 30 we is 3 40 plus we not divided by 3 40. Okay, so from here after solving we owe that is the spirit of the Observer comes out to be 32 m per second. So this become the answer for the part of the problem. Okay. Now solving for the part B in which we have to determine the wavelength of the sound that is linda S. Measured by the source for the part A. By the source. And for the second part it is for the observer. Okay and these two uh these two prevalence of the source will be seen. So hence we can see here linda S. It will be equal to the we divide by frequency, frequency of the source. Okay so we is here 3 40 m per second and frequency of the source is 5 30 her. So from here linda S. That is the wavelength of the source is equal to 0.64 m for the both parts. Okay thank you. Mhm.

Hello students. So in this question a source approaches a stationary observer. So observer has historic walls 20 m per second. And the speed of the source we as it is equals to 40 m per second. And the speed of the sound we it is given as 3 43 m per second and committing Earth sound of frequency. So source frequency is equal to $500. So we have to determine the observer frequency. Okay so that is observed frequency. So we know that if not it will be equals two Fs molecular baby. And here we notice zero. So we developed by and this V. S. It is the source is approaching, this is approaching towards the observer. Okay So hence we will get here we minus V. S. So substituting values so F. S. Which is 500 hertz Marie claire baby which is 3 43 m per second, divided by 3 43 m per second minus 40. Okay so from here after solving the observed frequency comes out to be 5 66 hertz, which can be nearly taken as 5 70. Hurt. So this becomes the answer for this problem. Okay. Okay. Thank you.

Hello students. So in this question we have given a sound of frequency 500 hertz. That is F. S. It is equal to 500 hertz. That is emitted by the stationary source of Yes, is equal to zero m per second here towards a receding observers. So we yes we owe. That is observer speed. It is receding. That is moving away. Okay. And we don't have the magnitude of the speed. Okay? Now uh the signal is reflected by the observer and received again by the source. Where the frequency is measured by observer and found two before 80 hertz. So we have to determine here the speed of the observer. Okay so when the sound is reflected back to the source from the observer. Okay So the observed frequency F. O. It is given by we minus V. O. D. V. Plus V. O. Molecular frequency of the source. Okay where we It is the speed of sound which is 3 43 m per second. So substituting the values. So we will get that F. O. Which is given as 4 80 hertz. This is equal to be which is 3 43 minus we owe developed by they were by 3 43 plus. We owe more clear by 500. Okay? So from here after solving we will get that we owe. It is equal to seven m per second. Okay? So this becomes the answer for the problem. Okay, So this is the the speed of the observer by which it is receding away. Okay? Thank you.

Hello students. So in this question a sound wave of frequency F. S. It is equal to 6 28 hertz is amateur by the stationary source, the speed of the source, it is equal to zero m per second towards an observer which is approaching with speed. V. O equals to 25 m per second. This speed is approaching speed. Okay, so we have to determine here the observed frequency by the observer. So for this problem the F. O. Will be equals two F. S. And we plus we owe the verb V. Okay, we is the speed of sound which is given in the question as 3 43 m per second. And here sources stationary. So we will get here only we and this is approaching so we will get assigned plus here. Okay, so substituting values so F. O. Will be equals to 6 28 herds Manipulated by V, which is 343 plus 25 divided by 343 meter per second. Okay, so from here, after solving observed frequency by the observer comes out to be 6 74 cards. So this becomes the answer for this problem. Okay, thank you.


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