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 laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

$5\ beats/second$ are heard when a tuning fork is sounded with sonometer wire under tension, when the length of the sonometer wire is either $0.95\ m$ or $1\ m$. The frequency of the fork will be:

  1. $195\ Hz$
  2. $150\ Hz$
  3. $300\ Hz$
  4. $251\ Hz$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When length is 0.95m

$v _1=\frac{v}{2\times 0.95}=\frac{v}{1.9}$
When length is 1m
$v _2=\frac{v}{2\times 1}=\frac{v}{2}$
$ v _1-v=5\quad v-v _2=5$
$ v _1-v _2=10$
$ \frac{v}{1.9}-\frac{v}{2}=10$
$\frac{0.1v}{3.8}=10$
$v=380m/s$
So, $v _1=200Hz , v _2=190Hz$
Then,
$v=195Hz$


Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

The length of a sonometer wire is $0.75\ m$ and density $9\times 10^3  k/m^3$It can bear a stress of $8.1\times 10^8 N/m^2$ with out exceeding the elastic limit The fundamental frequency that can be produced in the wire,is 

  1. $200\ Hz$
  2. $150\ Hz$
  3. $600\ Hz$
  4. $450\ Hz$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given, $Length=0.75m,density=9\times 10^3k/m^3,Stress=8.1\times10^8N/m^2$

Let the area of the wire be A.

So, $Stress=8.1\times10^8\Rightarrow Density=\dfrac{mass}{volume},mass=Density\times volume$

$=9\times10^3(0.75\times A)=9\times10^3 l\times A$ Where l is the length

$Mass=6.75\times10^3\times A\Rightarrow C=\sqrt{\dfrac{T}{mass/unit}}=\sqrt{\dfrac{8.1 \times 10^8(A)}{\dfrac{6.75\times10^3\times A}{0.75}}}=300m/s$

$f=\dfrac{c}{2l}=\dfrac{300}{1.5}=200Hz$
Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

The fundamental frequency in a stretched string is $100\space Hz$. To double the frequency, the tension in it must be changed to 

  1. $T _2 = 2T _1$
  2. $T _2 = 4T _1$
  3. $T _2 = T _1$
  4. $T _2 = \displaystyle\frac{T _1}{4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Fundamental frequency $\nu \propto \sqrt{T}$.
So, $\dfrac{\nu}{\nu'}=\sqrt{\dfrac{T}{T'}}\Rightarrow \dfrac{T}{T'}=\left(\dfrac{\nu}{\nu'}\right)^2=\dfrac{1}{4}$ 
$\Rightarrow T'=4T$. 

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

A sonometer wire supports a $4\ kg$ load and vibrates in fundamental mode with a tuning fork of frequency $426\ Hz.$ The length of the wire between the bridges is now doubled. In order to maintain fundamental mode, the load should be changed to 

  1. $1\ kg$
  2. $2\ kg$
  3. $8\ kg$
  4. $16\ kg$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Fundamental frequency f = (1/2L) * sqrt(T/m). If f is constant and L is doubled, sqrt(T) must double, meaning T must increase by a factor of 4. Since T is proportional to the load, the load must be 4 * 4kg = 16kg.

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

The density of the material of a wire used in sonometer is $7.5 \times 10 ^ { 5 } \mathrm { kg } / \mathrm { m } ^ { 3 }$  If the stress on the wire is $3.0 \times 10 ^ { 8 } \mathrm { N } / \mathrm { m } ^ { 2 }$ the speed of transverse wave in the wire will be-

  1. $100$ $\mathrm { m } / \mathrm { s }$
  2. $20$ $m / s$
  3. $300$ $m / s$
  4. $400$ $m / s$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

The total mass of a sonometer wire remains constant. On increasing the distance between two bridges to four times, its frequency will become

  1. $0.25\space times$
  2. $0.5\space times$
  3. $4\space times$
  4. $2\space times$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$f=\dfrac{1}{2L} \sqrt{\dfrac{T}{m}}$
if $L'=4L$
$f'=\dfrac{1}{8L} \sqrt{\dfrac{T}{m}}=\dfrac{1}{4}\dfrac{1}{2L} \sqrt{\dfrac{T}{m}}=\dfrac{1}{4} f$
Option "A" is correct.

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

If we add $8\space kg$ load to the hanger of a sonometer. The fundamental frequency becomes three times of its initial value. The initial load in the hanger was about 

  1. $4\space kg$
  2. $2\space kg$
  3. $1\space kg$
  4. $0.5\space kg$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$V=\sqrt { \dfrac { TL }{ m }  }$

$T:$tension

$m:$string mass

$L:$string length

$f=\dfrac { V }{ 2L } $          ----- fundamental frequency

suppose initial mass hanging is $M.$

$T=Mg$

${ f } _{ 1 }=\dfrac { V }{ 2L } $

${ f } _{ 1 }=\dfrac { 1 }{ 2L } \sqrt { \dfrac { MgL }{ m }  } $

$final \ \  mass=(M+8)$

${ f } _{ 2 }=\dfrac { 1 }{ 2L } $$\sqrt { \dfrac { (M+8)gL }{ m }  } $

 ${ f } _{ 2 }={ 3f } _{ 1 }$

 

On solving, we get

$M=1$ kg

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

A sonometer wire is to be divided in to three segments having fundamental frequencies in the ratio $1:2:3$. What should be the ratio of lengths?

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

$f=\dfrac{1}{2L} \sqrt{\dfrac{T}{m}}$
i.e. , for same sonometer, for different segments of bridges length will be the tuning parameter.
 we are given, 
$f _1:f _2:f _3=1:2:3$
while,
$f _1:f _2:f _3=\dfrac{1}{L _1}:\dfrac{1}{L _2}:\dfrac{1}{L _3}$
therefore $L _1:L _2:L _3=\dfrac{1}{f _1}:\dfrac{1}{f _2}:\dfrac{1}{f _3}=6:3:2$

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

The length of strings of a cello is $0.8\space m$. In order to change the pitch in frequency ratio $5/4$, their length should be decreased by

  1. $0.08\space m$
  2. $0.02\space m$
  3. $0.13\space m$
  4. $0.16\space m$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Frequency $\nu \propto \dfrac{1}{l}$. So, $\dfrac{\nu}{\nu'}=\dfrac{l'}{l}$
$\Rightarrow 1-\dfrac{l'}{l}=1-\dfrac{\nu}{\nu'}=1-\dfrac{4}{5}=0.2$
$\Rightarrow \Delta l=0.2\times l =0.16m$ 

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

Four wires of identical lengths, diameters and materials are stretched on a sonometer wire. The ratio of their tensions is 1 : 4 : 9 : 16. then, the ratio of their fundamental frequencies is 

  1. 1 : 4 : 9 : 16

  2. 1 : 2 : 3 : 4

  3. 16 : 9 : 4 : 1

  4. 4 : 3 : 2 : 1
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Fundamental frequency on the wire, $\nu = \dfrac{1}{2  L} \sqrt{\dfrac{T}{\mu}}$   

where, $\mu,   T$ and $L$ are the mass per unit length,  tension and length of the wire respectively.
Now for identical lengths, diameter and materials, $\nu \propto  \sqrt{T}$
Thus, $\nu _1  :  \nu _2  :  \nu _3 :  \nu _4   =  \sqrt{T _1}  :  \sqrt{T _2}  :  \sqrt{T _3}   :  \sqrt{T _4}$
  $\nu _1  :  \nu _2  :  \nu _3 :  \nu _4   =  \sqrt{1}  :  \sqrt{4}  :  \sqrt{9}   :  \sqrt{16}$
 $\implies      \nu _1  :  \nu _2  :  \nu _3 :  \nu _4   =  1  :  2  :  3   :  4$

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

A is point on a sonometer wire of uniform area and length L, such that the distances of A from the left end of the wire is $\dfrac { L }{ 18 } $ Find the amplitudes of vibration of the points A if the wire is set vibrating with maximum amplitude h in its ${ 3 }^{ rd }$ harmonic.

  1. 0.3 h

  2. 0.8 h

  3. 0.68 h

  4. 0.5 h
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

Four wires of identical lengths, diameters and of the same material are stretched on sonometer wire. The ratio of their tensions is 1 : 4 : 9 : 16. The ratio of their fundamental frequencies is

  1. 1:2:3:4

  2. 16:9:4:1

  3. 1:4:9:16

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

Frequency f is proportional to sqrt(T). If tensions are in ratio 1:4:9:16, frequencies are in ratio sqrt(1):sqrt(4):sqrt(9):sqrt(16), which is 1:2:3:4.

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

A sonometer wire of length l vibrates in fundamental mode when excited by a tunning fork of frequency 416 Hz. If the length is doubled keeping other things same, the string will

  1. vibrates with frequency of 416 Hz

  2. vibrates with frequency of 208 Hz

  3. vibrates with frequency of 832 Hz

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

since the wire is being excited by the tunning fork of frequency $ 416 $ Hz then wire of sonometer will always vibrate at frequency $ 416 $Hz the change in length will only effect its fundamental frequency. 

so the answer is A. 



Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

A sonometer wire under a tension of 10 kg weight is in unison with tuning fork of frequency 320 Hz. To make the wire vibrate in unison with a tuning fork of frequency 256 Hz, the tension should be altered by 

  1. 3.6 kg decreased

  2. 3.6 kg increased

  3. 6.4 kg decreased

  4. 6.4 kg increased
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

frequency in a sonometer is given as $f= \dfrac{v}{2l}\sqrt{\dfrac{T}{\mu}}$


$\dfrac{f _1}{f _2} = \sqrt{\dfrac{T _1}{T _2}}$

$\dfrac{320}{256} = \sqrt{\dfrac{10\times g}{T _2}}$

$\dfrac{5}{4} = \sqrt{\dfrac{10\times g}{T _2}}$

$T _2= \dfrac{16}{25} \times 10 g \ N$

$T _2 = 6.4 kg$

Tension to be decreased by 3.6 kg. 

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

A sonometer wire, with a suspended mass of $M=1 kg$, is in resonance with a given tuning fork. The apparatus is taken to the moon where the acceleration due to gravity is $\dfrac 16$ that on earth. To obtain resonance on the moon, the value of $M$ should be

  1. $1 kg$
  2. $\sqrt{6}$ kg
  3. $6 kg$
  4. $36 kg$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$f \propto \sqrt{T}=\sqrt{mg}$
$\therefore m _{1}g _{1}=m _{2}g _{2}$
$\Rightarrow (1)g=m\left ( \dfrac{g}{6} \right )$
$\Rightarrow m=6kg$