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 superposition and interference of sound waves

Consider ten identical sources of sound all giving the same frequency but having phase angles which are random. If the average intensity of each source is $I _{0}$, the average of resultant intensity $I$ due to all these ten sources will be

  1. $I = 100\ I _{0}$
  2. $I = 10\ I _{0}$
  3. $I = I _{0}$
  4. $I = \surd {10}\ I _{0}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For incoherent sources (random phases), the intensities add linearly. If there are 10 sources each with intensity I0, the total intensity is 10 * I0.

Multiple choice physics superposition and interference of sound waves

Two plane harmonic sound waves travelling in the same direction are given by the following displacement equations
$y _{1} (x, t) = A\cos (0.5\pi x - 100\pi t)$
$y _{2} (x, 1) = A\cos (0.46\pi x - 92\pi t)$
How may times, a listener can hear sound of maximum intensity in one second?

  1. $8$
  2. $6$
  3. $4$
  4. $3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The beat frequency is the difference between the two frequencies. f1 = 100/2 = 50 Hz, f2 = 92/2 = 46 Hz. Beat frequency = 50 - 46 = 4 Hz.

Multiple choice physics superposition and interference of sound waves

Beats are produced because of the superposition of two progressive notes> Maximum loudness at the waxing is $n$ times the loudness of either notes. What is the values of $n$?

  1. $4$
  2. $2$
  3. $\sqrt2$
  4. $1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Resultant amplitude is $A=\sqrt{A^2 _1+A^2 _2+2A _1A _2\cos(\phi)}$, where 


$\phi$ is the angle between the superposing waves.

$A=A _{max}=A _1+A _2$, when $\cos(\phi)=1$.

So, maximum loudness $I _{max}=A^2 _{max}=(A _1+A _2)^2=4A^2 _0=4I _0$, 

assuming that the waves have same amplitude $A _0$.

Multiple choice physics superposition and interference of sound waves

If a tuning fork sends a wave $5 sin \displaystyle \left(600\omega t - \frac{\pi}{0.6}x \right)$, then the amplitude of the intensity heard is

  1. $5$
  2. $5\sqrt{2}$
  3. $5\sqrt{3}$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\displaystyle \Delta x = 0.4m$
 $\displaystyle\Phi = k\Delta x = \frac{1\pi}{3} = \sqrt{5^2+5^2+2\times 5 cos  (2\pi/3)}$
$\displaystyle= 5.$

Multiple choice physics superposition and interference of sound waves

Two identical sources of sound of same frequency and identical intensities $\displaystyle I _0$ are producing sound. If their phases are irregular, then the average intensity of sound at a point where waves from the two sources are superposing is 

  1. $\displaystyle I _0$
  2. $\displaystyle 2 I _0$
  3. $\displaystyle 4I _0$

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

Given :   $I _1 = I _2  = I _o$


Resultant intensity        $I = I _1 + I _2 + 2  \sqrt{I _1  I _2}   cos \delta   $         
where $\delta $ is the phase difference.

 $I = I _o + I _o + 2  \sqrt{I _o \times  I _o}   cos \delta   = 2 (1 + cos  \delta)   $  

$\implies  I = 4  I _o  cos^2 \dfrac{\delta}{2}$

Now average intensity       $< I > = 4I _o  <cos^2  \dfrac{\delta}{2}>$

 $< I > = 4I _o  \times \dfrac{1}{2}  =  2  I _o$

Multiple choice physics superposition and interference of sound waves

A loudspeaker that produces signals from $50Hz$ to $500Hz$ is placed at the open end of a closed tube of length $1.1m.$ If velocity of sound is $330m/s,$ then frequencies that excites resonance in the tube are:

  1. $75\,Hz$
  2. $150\,Hz$
  3. $200\,Hz$
  4. $300\,Hz$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
If the length of the tube is $l$
then $l=\dfrac{x}{4}\Rightarrow \lambda =4l$
where $\lambda$ is the wavelength of the sound wave inside the tube If corresponding frequency is $V$ than 
$V\lambda=V$
$\therefore V=\dfrac{V}{\lambda}=\dfrac{V}{4l}=\dfrac{330}{4.(1.1)}$
$\therefore V=75\ Hz$
Therefore the fundamental tone freq of the tube is $V=75\ Hz$
This frequencies of the overtones are $3v, 5v, 3v .... $
i.e. $225, 373, 525$
If the loud speaker produces signals from $0\ Hz$ to $500\ Hz$ then frequencies that excites resonance in the tube are
$75\ Hz, 225\ Hz, 375\ Hz$
Multiple choice standing waves waves physics

A string attached to a tuning fork of frequency 300 Hz is made to vibrate. The other end of the string is fixed to a wall. If stationary waves are to be set up, what should be the phase of the reflected wave

  1. $\pi $ rads
  2. $\pi/2 $ rads
  3. $\pi/3 $ rads

  4. none of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The reflected and the actual wave has to be out of phase and only in such cases, a stationary wave is being setup

The correct option is (a)

Multiple choice standing waves waves physics

An organ pipe of length $80\ cm$ is opened at $x=0$ and closed at $x=80\ cm$. Speed of sound in the air column is $320\ m/sec$. If standing waves are generated in the closed organ pipe, then the correct equation of standing waves is/are (Here $s=$ longitudinal displacement, $P _{ex}=$ pressure excess) (Neglect the end correction).

  1. $S=A\cos\left(\dfrac{5\pi}{4}x\right)\sin\left(400\pi t\right)$
  2. $S=A\cos\left(\dfrac{5\pi}{8}x\right)\cos\left(1000\pi t\right)$
  3. $P _{ex}=A\cos\left(\dfrac{5\pi}{8}x\right)\sin\left(200\pi t\right)$
  4. $P _{ex}=A\sin\left(\dfrac{25\pi}{8}x\right)\cos\left(1000\pi t\right)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a pipe closed at one end (x=L) and open at the other (x=0), the pressure excess P_ex is zero at the open end (node) and maximum at the closed end (antinode). The standing wave equation for pressure is P_ex = A cos(kx) sin(wt). Given L=0.8m and v=320m/s, the fundamental frequency is v/4L = 100 Hz, so w = 200 pi. k = w/v = 200 pi / 320 = 5 pi / 8.

Multiple choice standing waves waves physics

A $string$ is stretched between fixed points separated by $75.0\ cm$. It is observed to have resonant frequencies of $420\ Hz$ and $315\ Hz$. There are no other resonant frequencies between these two.
Then, the lowest resonant frequency for this string is :

  1. $1.05$Hz
  2. $1050$Hz
  3. $10.5$Hz
  4. $105$Hz
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given,  $\dfrac { nV }{ 2L } =315\quad \longrightarrow (1)$

     &     $\dfrac { \left( n+1 \right) V }{ 2L } =420\quad \longrightarrow (2)$
equation (2) $-$ equation (1), we get
$\dfrac { \left( n+1 \right) V }{ 2L } -\dfrac { nV }{ 2L } =420-315$
$\Rightarrow \quad \left[ \dfrac { V }{ 2L } =105\quad { H } _{ 3 } \right] \rightarrow $  Lowest possible resonant frequency

$\therefore $  Option (D) is correct.

Multiple choice physics free, damped and forced oscillations damped harmonic motion damped oscillation free, forced and damped oscillations

The frequency of vibration is less than the natural frequency in

  1. Forced vibrations

  2. Free vibration

  3. Damped vibrations

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

It is our common experience that when a body is made to vibrate in a medium , the amplitude of the vibrating body continuously decreases with time and ultimately the body stops vibrating. This is called the damped vibrations.

Multiple choice physics beats in sound waves

What happens when two sound waves of frequencies differing by more than 10 Hz reach our ear simultaneously?

  1. beats are not produced

  2. the waves destroy each other's effect

  3. interference of sound does not take place.

  4. beats are produced but cannot be heard
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

 if the beat formation rate is too fast (more than 10 Hz) human ear could not distinguish the time interval of the two consicutive beats. 

Multiple choice physics beats in sound waves

You are listening to an $A$ note played on a violin string. Let the subscript $s$ refer to the violin string and $a$ refer to the air. Then:

  1. $f _{s}=f _{a}$ but $\lambda _{s}\neq\lambda _{a}$
  2. $f _{s}=f _{a}$ and $\lambda _{s}=\lambda _{a}$
  3. $\lambda _{s}=\lambda _{a}$ but $f _{s}\neq f _{a}$
  4. $\lambda _{s}\neq\lambda _{a}$ and $f _{s}\neq f _{a}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Frequencies of air and string is same because the source is same. Air and the string are two different medium so there velocities will also different. Hence the wavelength has to be different.

Multiple choice physics beats in sound waves

The fundamental frequency of a sitar wire is 440 vibrations per seond. If the sitar player reduces the length of the wire by $1/5^{th}$, then the change in the frequency of sitater wire will be

  1. 2200 vibrations/sec

  2. 1760 vibrations/sec

  3. 440 vibrations/sec

  4. 110 vibrations/sec
Reveal answer Fill a bubble to check yourself
A Correct answer