Question

A carousel that is 5.00 m in radius has a pair of 600-Hz sirens mounted on posts at opposite ends of a diameter. The carousel rotates with an angular velocity of 0.800 rad/s. A stationary listener is located at a distance from the carousel. The speed of sound is 350 m/s. The longest wavelength reaching the listener from the sirens is closest to

Answer 1

The answer is about 0,59 m or 59 cm the longest wavelength for a listener when the sirens is closest.

Step-by-step explanation:

Using the equation of the effect Doopler:

`f'= ((V + V_(L)) )/((V + V_(s)))*f_(o)`

Replacing:

W angular velocity = 0,8
`(rad)/(s)`

V = 350
`(m)/(s)`

`f_(o) = 600 Hz`

VL = 0
`(m)/(s)`
`( Is the velocity of the listener,  we assume he is at rest) </p><p>Vs = W*Radius ⇒ Vs = 0,8 [tex](rad)/(s)` * 5 m

Vs = 4
`(m)/(s)`

Replacing:

`f'= ((V + V_(L)) )/((V + V_(s)))*f_(o)`

`f'= (350 (m)/(s) )/((350 + 4) (m)/(s) ) * 600 Hz`

Note: Notice, the division have the same units so, it can be simplify

`f'= 593,220339 Hz`

Now using the frequency f' as the frequency of the listener to know L wavelength as the relation between f' and V

`L = (V)/(f')`

`L = (350 (m)/(s)  )/( 593,220339 Hz)`

Note: Remember Hz units are relations inverse of time so:
`Hz = (1)/(s)`

`L = (350 (m)/(s) )/( 593,220339 (1)/(s))`

`L = 0,59 m, or, 0,59 cm`