Tag: multiple reflection of sound and reverberation

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Multiple choice physics acoustics reverberation applications of reflection of sound multiple reflection of sound and reverberation

RADAR is used for

  1. locating submerged submarines

  2. receiving a signals in a radio receiver

  3. locating geostationary satellites

  4. detecting and locating the position of objects such as aeroplanes
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Radar is an acronym fro RAdio Detection And Ranging

$\rightarrow$ It is basically a detection system that uses radio waves to determine the range, angle or velocity of objects.
$\rightarrow$ More over; used for distance measurement
$\rightarrow$ Hence; can be used for detecting and locating the position of objects like aero planes

Multiple choice physics acoustics reverberation applications of reflection of sound multiple reflection of sound and reverberation

RADAR makes the use of.

  1. Infra-red waves of shorter wavelengths

  2. Radio-waves of very short wavelength

  3. Radio-waves of very long wavelength

  4. Ultraviolet waves of longer wavelengths
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

RADAR systems operate by transmitting radio waves of very short wavelengths (microwaves), which reflect off objects to detect their position and speed.

Multiple choice physics acoustics reverberation applications of reflection of sound multiple reflection of sound and reverberation

A rocket is moving at a speed of $220\,\,m\,\,s^{-1}$ towards a stationary target, emits a sound of frequency  $1000 Hz$. Some of the sound reaching the target gets reflected back to the rocket as echo. The frequency of the echo as detected by the rocket is
(Take velocity of sound $= 330\,\,m\,\,s^{-1}$)

  1. $3500 Hz$
  2. $4000 Hz$
  3. $5000 Hz$
  4. $3000 Hz$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

As the source (i.e rocket) is moving toward the stationary target, therefore the frequency of sound detected by the target is
$\upsilon' = \dfrac{\upsilon _0\nu}{\nu - \nu _s} = \dfrac{1000 \times 330}{330 - 220} = \dfrac{1000 \times 330}{110} = 3000 Hz$
Now the target is the source (as it is the source of echo) and the rocket's detector is the observer who intercepts the echo of frequency \acute{\upsilon}. Hence, the frequency of the echo detected by the rocket is


$\upsilon" = \dfrac{\upsilon'(\nu + \nu _O)}{\nu} = \dfrac{3000(330 + 220)}{330} = 5000 Hz $