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8Q

Expert-verifiedFound in: Page 505

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**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 **

** **

**Answer**

- The ranks of the curves according to the linear speedof the sound are

Curve 3 > Curve 2 = Curve 1.

- The ranks of the curves according to the angular speed
*ω*of the merry-go-rounds are Curve 1 > Curve 2 = Curve 1. - The ranks of the curves according to the radii of the merry-go-rounds are

Curve 3 > Curve 2 > Curve 1.

** **

** **

See Figure 17-29

**By using the formulae for the speed v of sound, the angular frequency ω**

The formulae are as follows:

- The speed of sound is

$v=f\lambda $

2.The angular frequency *ω* is

$\omega =\frac{2\pi}{T}$

3.The linear speed is

$v=r\omega $

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 3^{rd }merry-go is highest and it is same for the 1^{st} and the 2^{nd}. So, the linear velocity of the source on the 3^{rd }merry-go round would be higher and the same goes for the 1^{st} and the 2^{nd}.

** **

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 1^{st} one would have highest angular velocity, as the wave has the smallest time period out of all the waves. The angular velocity of the 2^{nd} and the 3^{rd} would be the same and smaller than that of the 1^{st.}, because they have the same time period which is greater than that of the 1^{st} wave. Therefore, we can write

1 > 2 =3

Linear velocity of the merry-go-round can be written as

$v=r\omega $

Therefore

${r}_{1}=\frac{{v}_{1}}{{\omega}_{1}}\phantom{\rule{0ex}{0ex}}{r}_{2}=\frac{{v}_{2}}{{\omega}_{2}}\phantom{\rule{0ex}{0ex}}{r}_{3}=\frac{{v}_{3}}{{\omega}_{3}}$

From the first two cases, it is clear that the linear velocity of the 1^{st} 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 3^{rd}.

The linear velocity of the 1^{st} and 2^{nd} 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 1^{st} is greater than the angular velocity of the 2^{nd}. Therefore, the radius of the 1^{st} would be smaller than that of the 2^{nd}. From this, the 3^{rd} has the largest radius, followed by the 2^{nd}, and the 1^{st} 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.

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