Question: A friend rides, in turn, the rims of three fast merry-go-roundswhile holding a sound source that emits isotropically at a certain frequency. You stand far from each merry-go-round. The frequency you hear for each of your friend’s three rides varies as the merry-go-round rotates. The variations in frequency for the three rides aregiven by the three curves in Fig. 17-29. Rank the curves according to:
(a) the linear speed v of the sound source, (b) the angular speeds v ofthe merry-go-rounds, and (c) the radii r of the merry-go-rounds, greatest first
Curve 3 > Curve 2 = Curve 1.
Curve 3 > Curve 2 > Curve 1.
See Figure 17-29
By using the formulae for the speed v of sound, the angular frequency ω and the linear speed v, find the ranks of the curves according to the linear speed v of the sound, according to the angular speed ω of the merry-go-rounds, and according to the radii r of the merry-go-rounds.
The formulae are as follows:
2.The angular frequency ω is
3.The linear speed is
where f is frequency, r is radius, ω is an angular frequency, λ is wavelength, T is time and v is velocity
From the Doppler Effect, the apparent frequency of the source of sound would be maximum when the velocity of the source approaching towards it is the maximum. As seen in the graph, the apparent frequency of sound for 3rd merry-go is highest and it is same for the 1st and the 2nd. So, the linear velocity of the source on the 3rd merry-go round would be higher and the same goes for the 1st and the 2nd.
The time taken by the merry-go-round to complete one rotation would be equivalent to the time period of one complete wave shown in the graph. As angular velocity is inversely proportional to the time period, we can conclude that the 1st one would have highest angular velocity, as the wave has the smallest time period out of all the waves. The angular velocity of the 2nd and the 3rd would be the same and smaller than that of the 1st., because they have the same time period which is greater than that of the 1st wave. Therefore, we can write
1 > 2 =3
Linear velocity of the merry-go-round can be written as
From the first two cases, it is clear that the linear velocity of the 1st is the greatest and the angular velocity is the smallest. Using this, it can be concluded that the radius would be the largest for the 3rd.
The linear velocity of the 1st and 2nd is the same. So, the radius value for these two would be inversely proportional to the angular velocity. Now, from the second case, the angular velocity of the 1st is greater than the angular velocity of the 2nd. Therefore, the radius of the 1st would be smaller than that of the 2nd. From this, the 3rd has the largest radius, followed by the 2nd, and the 1st would have the smallest radius; that is, 3>2>1.
Thus, by using the formulae for the speed of sound, the angular frequency ω and the linear speed , the answer to this question can be found.
Figure, shows two points sources and that emit sound of wavelength The emission are isotropic and in phase, and the separation between the sources is . At any point P on the , the wave from and the wave from interfere. When P is very away , what are (a) the phase difference between the arriving waves from and (b) the type of interference they produce? Now move point P along the towards. (c) Does the phase difference between the waves increase or decrease? At what distance do the waves have a phase difference of (d) 0.50 (e)1.00 (f)1.50?
A well with vertical sides and water at the bottom resonates at 7.00Hz and at no lower frequency. (The air-filled portion of the well acts as a tube with one closed end and one open end.) The air in the well has a density of 1.10kg/m3 and a bulk modulus of . How far down in the well is the water surface?
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