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 study of sound reverberation applications of reflection of sound multiple reflection of sound and reverberation applications of ultrasound

A pendulum has a frequency of 5 vibrations per second. An observer starts the pendulum and fire a gun simultaneously. He hears echo from the cliff after 8 vibrations of the pendulum. If the velocity of sound in air is 340  $ms^{-1}$, find the distance between the cliff and the observer :

  1. 252 m

  2. 240 m

  3. 272 m

  4. 182 m
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The sensation of any sound persists in our ear for about 0.1 seconds. This is known as the persistence of hearing. If the echo is heard within this time interval, the original sound and its echo cannot be distinguished. So the most important condition for hearing an echo is that the reflected sound should reach the ear only after a lapse of at least 0.1 second after the original sound dies off. 
It is given that a pendulum has a frequency of $5$ vibrations per second. An observer starts the pendulum and fire a gun simultaneously. He hears echo from the cliff after $8$ vibrations of the pendulum which means that he hears the echo after 1.6 seconds. The speed of sound is given as $340 m/s$.
The distance traveled by sound of the gun shot in $1.6 $ seconds is calculated from the formula 
Distance traveled $=velocity\quad of\quad sound\times time\quad taken$. That is, $340 \times 1.6 $ = 544 m. This is twice the minimum distance between a source of sound (man) and the reflector (cliff) as it is reflected sound. 
So, the cliff is at a distance of 272 m least from the man, for the reflected sound or the echo to be heard distinctly in 1.6 seconds.

Multiple choice physics study of sound reverberation applications of reflection of sound multiple reflection of sound and reverberation applications of ultrasound

The term reverberation time is generally understood to be the reverberation time at which of the following frequencies?

  1. 1024 Hz

  2. 2048 Hz

  3. 512 Hz

  4. 256Hz
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When a reflecting surface is less than 17 m away from the source, the echo that returns appears to be the original sound, just prolonged. This effect is called reverberation. The time interval between the original sound and the returning echo must be less than 0.1 seconds. It depends on the adsorption coefficient of the material which usually defines at 2048 Hz frequency.

Multiple choice physics sound sound as a wave of disturbance vibrations in a tuning fork vibrations in tuning fork

The frequency of a man's voice is $300\space Hz$. If the velocity of sound waves is $336\space ms^{-1}$, the wavelength of the sound is

  1. $1.12\space m$
  2. $300\times336\space m$
  3. $330/336\space m$

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

We are given, frequency $f=300 Hz$, and velocity $v=336 ms^{-1}$,

From the relation $v=f \times \lambda$,

$\lambda = v/f =336/300=1.12 m$

Option "A" is correct.

Multiple choice physics sound sound as a wave of disturbance vibrations in a tuning fork vibrations in tuning fork

A rod $70\space cm$ long is clamped from middle. The velocity of sound in the material of the rod is $3500\space ms^{-1}$. The frequency of fundamental note produced by it is :

  1. $3500\space Hz$
  2. $2500\space Hz$
  3. $1250\space Hz$
  4. $700\space Hz$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

SInce rod is clamped at middle, therefore only nodes can form at that point, and at free end only antinode can form, therefore for fundamental mode of frequency,

$\lambda/4=L/2$

$\lambda=2L=2\times 70 cm=1.4 m$

$v=3500\  ms^{-1}$ is given, 

We know, $v=f \times \lambda$ or $f = \dfrac{v}{\lambda}=3500/1.4=2500 Hz$

Option "B" is correct.

Multiple choice physics sound sound as a wave of disturbance vibrations in a tuning fork vibrations in tuning fork

A sonometer wire, $100\ \text{cm}$ in length has a fundamental frequency of $330\ \text{Hz}$. The velocity of propagation of transverse waves along this wire is :

  1. $330\ \text{ms}^{-1}$
  2. $660\ \text{ms}^{-1}$
  3. $115\ \text{ms}^{-1}$
  4. $990\ \text{ms}^{-1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For the fundamental frequency $(f _0=330 Hz),\ \lambda=2 L=200\ \text{cm} = 2\ \text{m}$


then from the formula, $v=f _0 \times \lambda= 330 \times 2.0= 660 \text{ms}^{-1}$

Multiple choice physics sound sound as a wave of disturbance vibrations in a tuning fork vibrations in tuning fork

If the frequency of a sound wave is increased by 25%, then the change in its wavelength will be

  1. 25% decrease

  2. 20% decrease

  3. 20% increase

  4. 25% increase
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
The frequency  of wave is given by

$\nu = \dfrac{v}{\lambda}$

$\lambda = \dfrac{v}{\nu}$

When the frequency of a sound wave is increased by 25%, then the new 
wavelength is

$\lambda' = \dfrac{v}{\nu+\dfrac{25}{100}\nu}$

$\lambda' = \dfrac{v}{\nu+\dfrac{1}{4}\nu}$

$\lambda' = \dfrac{v}{\dfrac{5}{4}\nu}$

$\lambda' = \dfrac{4v}{5\nu}$

Hence, the percent change in wavelength is

$\dfrac{\lambda - \lambda'}{\lambda} \times 100 = \dfrac{\dfrac{v}{\nu} - \dfrac{4v}{5\nu}}{\dfrac{v}{\nu}} \times 100$

$\Rightarrow -\dfrac{1}{4} \times 100 = -20$%.

Hence, wavelength decreases by 20%
Multiple choice physics sound sound as a wave of disturbance vibrations in a tuning fork vibrations in tuning fork

A $40\ cm$ long brass rod is dropped, one end first on to a hard floor but it is caught before it topples over. With an oscilloscope it is determined that the impact produces a $3\ kHz$ tone. The speed of sound in brass is:

  1. $1200\ m/s$
  2. $2400\ m/s$
  3. $3600\ m/s$
  4. $3000\ m/s$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Both ends are free and therefore antinodes are formed.
The relation between the wavelength of the wave and the length of the rod for fundamental frequency will be:

$\Rightarrow l=\dfrac{\lambda}{2} \ \ \Rightarrow \lambda=2l$

The speed of the wave in the rod is:
$v=f\lambda = 2\times 40\times 3\times 10^{3}$$=2400\ ms^{-1}$

Multiple choice physics sound sound as a wave of disturbance vibrations in a tuning fork vibrations in tuning fork

The frequency of a man's voice is 300 Hz and its wavelength is 1 meter. If the wavelength of a child's voice is 1.5 m, then the frequency of the child's voice is :

  1. 200 Hz

  2. 150 Hz

  3. 100 Hz

  4. 350 Hz.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\nu _1\lambda _1= \nu _1\lambda _1$ since $v= \nu\lambda$ is same for both a man and child.

$ \therefore 300 \times 1 =  \nu _2 \times  1.5$


$ \Rightarrow \nu _2 = 200 : Hz$

Multiple choice physics superposition of waves-1: interference and beats sound as a wave of disturbance vibrations in a tuning fork vibrations in tuning fork

Two sound waves of wavelengths $1\space m$ and $1.01\space m$ produce $10$ beats in $3$ seconds. Then, the velocity of the sound is

  1. $330\space ms^{-1}$
  2. $333.3\space ms^{-1}$
  3. $336.7\space ms^{-1}$

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

From the formula, $v=f \lambda$

and the velocity of the sound in air is frequency independent.

for two frequencies $f _1$ and $f _2$, wavelengths are $\lambda _1$ 

and $\lambda _2$ , then the beats will be

$f _1-f _2=v\left(\dfrac{1}{\lambda _1}-\dfrac{1}{\lambda _2}\right)=v\left(\dfrac{\lambda _2-\lambda _1}{\lambda _2  \lambda _1}\right)$

$\dfrac{10}{3}=v\left(\dfrac{0.01}{1.0\times 1.01}\right)$

$v=\dfrac{10\times 1.0\times 1.01}{3.0\times 0.01}=336.66\ \text{ms}^{-1}$

Option "C" is correct.

Multiple choice physics study of sound sound as a wave of disturbance vibrations in a tuning fork vibrations in tuning fork

The sound propagates in a gaseous medium as :

  1. transverse waves

  2. longitudinal waves

  3. both (a) and (b)

  4. neither (a) nor (b)
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Sound is a sequence of waves of pressure which propagates through compressible media such as air or water. Sound propagates as longitudinal waves in gaseous medium. Longitudinal waves are the waves in which the displacement of the medium is in the same direction, or the opposite direction to the direction to the direction of propagation of wave.