Physics · Science General

Acoustics and Sound Waves

2,160 Questions

Acoustics and sound waves deal with mechanical vibrations traveling through media like air and water. Key concepts include wave reflection, beats, echoes, the Mach number, and infrasound. These physics fundamentals are regularly tested in general science sections of multiple competitive exams.

Sound wave propagationEchoes and reflectionWave interferenceMach numberInfrasound frequency

Acoustics and Sound Waves Questions

Multiple choice physics medical imaging ultrasonic sound using ultrasound in medicine ultrasound and its applications

The properties of ultrasound that make it useful is/are 

  1. high power and high speed.

  2. high power and good directivity.

  3. high frequency and high speed.

  4. high frequency and bending around the objects.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Answer is B.
The ultrasonic devices achieve high directivity by modulating audible sound onto high frequency ultrasound. The higher frequency sound waves have a shorter wavelength and thus don't spread out as rapidly. For this reason, the resulting directivity of these devices is far higher than physically possible with any loudspeaker system. However, they are reported to have limited low-frequency reproduction abilities. 
High power ultrasound can break up stony deposits or tissue, accelerate the effect of drugs in a targeted area, assist in the measurement of the elastic properties of tissue, and can be used to sort cells or small particles for research.
Hence, the above characteristics make ultrasound to be very useful.

Multiple choice physics medical imaging ultrasonic sound using ultrasound in medicine ultrasound and its applications

The ultrasound waves have a much greater penetrating power than ordinary sound because they have very high

  1. Amplitude

  2. Frequency

  3. Wave length

  4. Speed
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The frequency $f$ is related with intensity $I$ of wave by ,

           $I\propto f^{2}$ ,
and intensity is related with energy $E$ by ,
          $E\propto I$ ,
therefore due to high frequency ultrasound has a greater energy than ordinary sound wave , hence greater penetrating power because penetrating power depends upon energy of the wave .

Multiple choice physics medical imaging ultrasonic sound using ultrasound in medicine ultrasound and its applications

Mark the correct statement:

  1. Human beings cannot hear ultrasound.

  2. Dogs, bats and dolphins can hear ultrasound.

  3. Ultrasound have short wavelength.

  4. All
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The sound waves having frequency higher than 20,000 Hz is

called ultrasonic waves. Human beings can"t hear Ultra sound. Bats and dolphins detect the presence of any obstacle by

hearing the echo of the sound produced by them.

Multiple choice physics superposition of waves-2: stationary (standing) waves: vibrations of air columns from moving to stationary stationary (or standing) waves formation of stationary waves

As two students holds opposite ends of slinky that is the resting on the floor, one student shakes the end he is holding back and forth with the constant frequency. He later shakes exactly the same way but with a much greater frequency.
Which statement best describes other changes that take place as a result of this increased frequency?

  1. The wave speed and the wavelength both increases.

  2. The wave speed increases, but the wavelength does not significantly change.

  3. The wavelength increases ,but the wave speed does not significantly change.

  4. The wave speed decreases, but the wavelength does not significantly change.

  5. The wavelength decreases ,but the wave speed does not significantly change.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We know that frequency of a wave is characterized by the source of wave . In both the cases wave in the slinky has a constant frequency but not same in both cases  , 

now we have $v=f\lambda$ ,
or                  $f=v/\lambda$ , 
when frequency $f$ is  increased in second case , wave speed $v$ increases and to maintain a constant frequency $f$ (to maintain a constant ratio), wavelength $\lambda$ also increases
 .

Multiple choice physics oscillations and waves standing waves in strings standing waves reflection of waves

When we hear a sound, we can identify its source from : 

  1. Amplitude of sound

  2. Intensity of sound

  3. Wavelength of sound

  4. Overtones present in the sound
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Answer is D.

When we hear a sound, we can identify its source from overtones present in the sound.
The fundamental is the frequency at which the entire wave vibrates. Overtones are other sinusoidal components present at frequencies above the fundamental. All of the frequency components that make up the total waveform, including the fundamental and the overtones, are called partials. Together they form the harmonic series.
Overtones which are perfect integer multiples of the fundamental are called harmonics. When an overtone is near to being harmonic, but not exact, it is sometimes called a harmonic partial, although they are often referred to simply as harmonics. Sometimes overtones are created that are not anywhere near a harmonic, and are just called partials or inharmonic overtones.

Multiple choice physics oscillations and waves standing waves in strings standing waves reflection of waves

A set of 3 standing waves 5, 10 and 15 Hz are to be setup on a string fixed at one end. One of these frequencies are suppressed, while passing through it. Identify them:

  1. 5 Hz

  2. 10 Hz

  3. 15 Hz

  4. All the frequencies will pass through them
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In a string fixed at one end, only odd harmonics are allowed and even harmonics are suppressed.

Thus the 10Hz standing wave is suppressed,
The correct option is (b)

Multiple choice physics oscillations and waves standing waves in strings standing waves reflection of waves

Find the number of beats produced per sec by the vibrations $x _1=A\sin (320\pi t)$ and $x _2=A\sin (326\pi t)$.

  1. 3

  2. 4

  3. 5

  4. 6
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$X _1=Asin(320\Pi t)$

$X _2=Asin(326\Pi t)$
On comparing it with general equation.
$X=Asin(wt)$
Then, $w _1=320\Pi $
$w _1=2\Pi f$
frequency=160 Hz
Similarly,
$w _2=326\Pi $
$w _2=2\Pi f$
frequency=163 Hz
No of beats=163-160=3

Multiple choice physics oscillations and waves standing waves in strings standing waves reflection of waves

In an organ pipe(may be closed or open) of $99$ cm length standing wave is setup, whose equation is given by longitudinal displacement.
$\xi =(0.1mm)\cos \dfrac{2\pi}{0.8}(y+1cm)\cos 2\pi (400)t$
where y is measured from the top of the tube in meters and t is second. Here $1$cm is the end correction.
The air column is vibrating in :

  1. First overtone

  2. Fifth harmonic

  3. Third harmonic

  4. Fundamental mode
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation is of the form cos(k(y+c))cos(wt). The angular frequency w = 2*pi*400, so f = 400 Hz. The wave number k = 2*pi/0.8, so wavelength lambda = 0.8 m = 80 cm. With end correction, the effective length is 99 + 1 = 100 cm. For a closed pipe, L = (2n-1)lambda/4. 100 = (2n-1)80/4 = (2n-1)20. 5 = 2n-1, so n=3. This corresponds to the 5th harmonic.

Multiple choice physics oscillations and waves standing waves in strings standing waves reflection of waves

In a reasonance tube experiment, a closed organ pipe of lenght $120$ cm is used. initially it is completely fiiled with water. It is vibrated with tuning fork of frequency $340$ Hz. To achieve reasonance the water level is lowered then (given ${V _{air}} = 340m/\sec $., neglect end correction):

  1. minimum lenght of water column to have the resonance is 45 cm.

  2. the distance between two successive nodes is 50 cm.

  3. the maximum lenght of water column to resonance is 95 cm.

  4. the distance between two successive nodes is 25 cm.
Reveal answer Fill a bubble to check yourself
D Correct answer