Physics · Science General

Acoustics and Sound Waves

2,006 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 stationary waves determining wavelength and speed of sound resonance tube resonance and sonometer

The frequency of a whistle is 200 Hz. It is approaching to stationary observer with a speed 1/3 the speed of sound. The frequency of sound as heard by the observer will be 

  1. $450 Hz$
  2. $300 Hz$
  3. $400 Hz$
  4. $425 Hz$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the Doppler effect formula for a source approaching a stationary observer: f' = f * (v / (v - vs)). Given vs = v/3, f' = 200 * (v / (v - v/3)) = 200 * (v / (2v/3)) = 200 * (3/2) = 300 Hz.

Multiple choice physics stationary waves determining wavelength and speed of sound resonance tube resonance and sonometer

The amplitude of vibration of the particles of air through which a sound wave of intensity $2.0 \times 10 ^ { - 6 } \mathrm { Wm } ^ { - 2 }$ and frequency $1.0 kHz$ is passing - (Density of air = 1.2 $k g m ^ { - 3 }$  and speed of sound in air = 330 $m s ^ { - 1 }$ is)

  1. $4.4 \times 10 ^ { - 8 } m$
  2. $1.6 \times 10 ^ { - 8 } m$
  3. $2.4 \times 10 ^ { - 6 } m$
  4. $1.8 \times 10 ^ { - 6 } m$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The intensity of a sound wave is given by I = 2 * pi^2 * f^2 * A^2 * rho * v. Rearranging for amplitude A: A = sqrt(I / (2 * pi^2 * f^2 * rho * v)). Plugging in values: I = 2e-6, f = 1000, rho = 1.2, v = 330. A = sqrt(2e-6 / (2 * 9.87 * 1e6 * 1.2 * 330)) = 1.6e-8 m.

Multiple choice physics stationary waves determining wavelength and speed of sound resonance tube resonance and sonometer

The longitudinal waves travel in a coiled spring at a rate of 10 m/s. The distance between two consecutive compressions is 25cm. What is the frequency of the waves?

  1. 25Hz

  2. 10Hz

  3. 40Hz

  4. 250Hz
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Answer is C.

A sound wave has a speed that is mathematically related to the frequency and the wavelength of the wave. The mathematical relationship between speed, frequency and wavelength is given by the following equation.
Speed = Wavelength * Frequency. That is, Frequency = Speed / Wavelength.
In this case, the frequency is 140 per second and wavelength is 25 cm, that is, 0.25 m.
Therefore, Frequency = 10 / 0.25  = 40 Hz.
The frequency of the wave is 40 Hz.

Multiple choice physics stationary waves determining wavelength and speed of sound resonance tube resonance and sonometer

A hospital uses an ultrasonic scanner to locate tumours in a tissue. The operating frequency of the scanner is $4.2$ $MH _z$. The speed of sound  in a tissue is $1.7$ ${km/s}$. The wavelength of sound in tissue is close to

  1. $4\times 10^{-4}$ $m$
  2. $8\times 10^{-4}$ $m$
  3. $4\times 10^{-3}$ $m$
  4. $8\times 10^{-3}$ $m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given:
Frequency $(f)=4.2$ $MH _z = 4.2\times 10^{6}$ $H _z$
Speed in tissue $(v)=1.7$ ${km/s} = 1700$ ${m/s}$
$\therefore$ Wavelength $=\lambda \times f=v$
$\lambda=\cfrac{v}{f}=\cfrac{1700}{4.2\times 10^{6}}=4\times 10^{-4}$ $m$

Multiple choice physics superposition of waves-2: stationary (standing) waves: vibrations of air columns determining wavelength and speed of sound resonance tube resonance and sonometer

A resonance tube apparatus is employed to.

  1. Investigate the dependence of velocity of sound in air upon temperature

  2. Verify the laws of vibrating strings

  3. Study beats

  4. Determine the velocity of sound in air
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A resonance tube is a classic laboratory apparatus used to determine the speed of sound in air by measuring the lengths of air columns that resonate with a tuning fork of known frequency.

Multiple choice physics stationary waves determining wavelength and speed of sound resonance tube resonance and sonometer

Let ${ n } _{ 1 }$ and ${ n } _{ 2}$ be the two slightly different frequencies of two sound waves. The time interval between waxing and immediate next waning is ..........

  1. $\cfrac { 1 }{ { n } _{ 1 }-{ n } _{ 2 } } $
  2. $\cfrac { 2 }{ { n } _{ 1 }-{ n } _{ 2 } } $
  3. $\cfrac { { n } _{ 1 }-{ n } _{ 2 } }{ 2 } $
  4. $\cfrac { 1 }{ { 2(n } _{ 1 }-{ n } _{ 2 }) } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Beat frequency during constructive interference(waxing) is ($n _1-n _2$)
Beat frequency during destructive interference (waning) is ($n _1-n _2$)
The combination of two waves will give beat frequency as $2(n _1-n _2)$
Now ,the number of beats produced per one second is defined as the reciprocal of difference in frequencies two sound waves which produce waxing and waning.
$\therefore\ $ Time interval between waxing and immediate waning is $=\dfrac{1}{2(n _1-n _2)}$ 
Multiple choice physics superposition of waves-2: stationary (standing) waves: vibrations of air columns determining wavelength and speed of sound resonance tube resonance and sonometer

In a resonating air column, the first booming sound is heard when the length of air column is $10\ cm$. The second booming sound will be heard when length is:

  1. $20\ cm$
  2. $30\ cm$
  3. $40\ cm$

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Booming sound indicates that at that length, $l _1$, the air column is in resonance with the given frequency.
and that length is,
$l _1= \lambda /4=10$
or, $\lambda = 40cm$
The next resonance length will be :
$l _2=3\lambda/4=30 cm$

Multiple choice physics stationary waves determining wavelength and speed of sound resonance tube resonance and sonometer

In Kundt's tube, when waves of frequency $10^3\space Hz$ are produces the distance between five consecutive nodes is $82.5\space cm$. The speed of sound in gas filled in the tube will be

  1. $660\space ms^{-1}$
  2. $330\space ms^{-1}$
  3. $230\space ms^{-1}$
  4. $100\space ms^{-1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\quad \displaystyle\frac{5\lambda}{2} = 82.5\space cm$

$\quad \lambda = 33\space cm\quad and \quad v = f\lambda = 330\space ms^{-1}$ 

Multiple choice physics stationary waves determining wavelength and speed of sound resonance tube resonance and sonometer

The frequency of a fork is $500$Hz. Velocity of sound in air is $350$ $ms^{-1}$. The distance through which sound travel by the time the fork makes $125$ vibrations is?

  1. $87.5$m
  2. $700$m
  3. $1400$m
  4. $1.75$m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$wavelength=\dfrac { velocity }{ frequency } $ 

$=\dfrac { 350 }{ 500 } =\dfrac { 7 }{ 10 } $
Distance traveled in $125$ vibrations
$=$wavelength$\times$ no of vibrations
$=\dfrac { 7 }{ 10 } \times 125$
 $=87.15m$

Multiple choice physics stationary waves determining wavelength and speed of sound resonance tube resonance and sonometer

Frequency of tuning fork $A$ is $256\ Hz.$ It produces four beats/sec with tuning fork $B.$ When wax is applied at tuning fork $B$ then $6$ beats/sec are heard. By reducing little amount of wax $4$ beats/sec are heard. Frequency of $B$ is : 

  1. $250\ Hz$
  2. $252\ Hz$
  3. $260\ Hz$
  4. $256\ Hz$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the unknown frequency of the tuning fork be x.

So, according to the given data when no waxed, its frequency must be,

$x=256\pm 4$  to produced a beat of $4\ beats /sec$.

We know, the frequency of a tuning fork decreases as it is waxed.

So, to produce $6\  beats/s$, after being waxed, the frequency of the tuning fork must be

  $ x=256-4 $

 $ x=252\,Hz $

Hence, the frequency of $B$ is $252\ Hz$

 

Multiple choice physics superposition of waves-2: stationary (standing) waves: vibrations of air columns determining wavelength and speed of sound resonance tube resonance and sonometer

In a resonace air column experiment, first and second resonance are obtained at length of air columns $l _{1}$ and $l _{2}$ the third resonance will be obtained at a length of

  1. $2l _{2}-l _{1}$
  2. $l _{2}-2l _{1}$
  3. $l _{2}-l _{1}$
  4. $3l _{2}-l _{1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In a resonance tube, the resonance lengths are l1 = lambda/4, l2 = 3*lambda/4, and l3 = 5*lambda/4. The difference between consecutive resonances is lambda/2. Thus, l3 = l2 + (l2 - l1) = 2*l2 - l1.

Multiple choice physics stationary waves determining wavelength and speed of sound resonance tube resonance and sonometer

A person observes a change of 2.5% in frequency of sound of horn of a car . If the car is apporaching forward the person  sound velocity is 320 m/s then velocity of car in m/ s wil be appromately

  1. 8

  2. 800

  3. 7

  4. 6
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Doppler formula n'$=\dfrac{nv}{v-v _s}$ $n' > n$
if $n2100\quad n'=102.5$
Since source is moving towards distance so
$102.5=\dfrac{100\times 320}{320-v _s}$
$\therefore v _s=8$m/sec.
Multiple choice physics stationary waves determining wavelength and speed of sound resonance tube resonance and sonometer

In an experimental determination of the velocity of sound using a Kundt's tube, standing waves are set up in the metallic rod as well as in the rigid tube containing air, both the waves have the same :

  1. amplitude

  2. frequency

  3. wavelength

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

Speed, wavelength and amplitude change as it is traveling through different material on the other side frequency must remain constant to conserve energy (which is dependent solely on frequency).