0

Determining wavelength and speed of sound - class-IX

Description: determining wavelength and speed of sound
Number of Questions: 35
Created by:
Tags: wave motion physics superposition of waves-2: stationary (standing) waves: vibrations of air columns stationary waves sound waves
Attempted 0/35 Correct 0 Score 0

As an empty vessel is filled with water, its fundamental frequency

  1. Increases

  2. Decreases

  3. Remains the same

  4. None of these


Correct Option: A
Explanation:

   An empty vessel with a base is like a closed organ pipe , and fundamental frequency of a closed organ pipe is given by ,

       $n _{1}=v/4l$ ,
where $l=$ length of air column in pipe (height of pipe) ,
  from above we get,    
       $n _{1}\propto 1/l$ ,
when the empty vessel is filled with water , the length of air column $l$ in pipe decreases and therefore fundamental frequency $n _{1}$ increases as length of air column and fundamental frequency are inversely proportional to each other .

In Kundt's tube experiment the metallic rod executes 

  1. transverse vibration.

  2. longitudinal vibrations.

  3. both.

  4. none of these


Correct Option: B
Explanation:

In metallic rod longitudinal waves travels in vibrations.
Option "B" is correct. 

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. $425 Hz$


Correct Option: B

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. $1.8 \times 10 ^ { - 6 } m$


Correct Option: B

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


Correct Option: C
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.

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. $8\times 10^{-3}$ $m$


Correct Option: A
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 $=\lambda \times f=v$
$\lambda=\cfrac{v}{f}=\cfrac{1700}{4.2\times 10^{6}}=4\times 10^{-4}$ $m$

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


Correct Option: D

The audible range of frequency of sound waves for human beings is _____.

  1. 10 Hz to 10,000 Hz

  2. 20 Hz to 20,000 Hz

  3. 5 Hz to 50,000 Hz

  4. 50 Hz to 20,000 Hz


Correct Option: B

If the frequency of human heart is $1.25$ Hz, the number of heart beats in $1$ minute is

  1. $65$

  2. $75$

  3. $80$

  4. $90$


Correct Option: B
Explanation:

Beat frequency of heart = $1.25$Hz.
$\therefore$ Number of beats in $1$ minute = $1.25 \times 60 = 75$.

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 }) } $


Correct Option: D
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 $=\dfrac{1}{2(n _1-n _2)}$ 

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


Correct Option: B
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 :
$l _2=3\lambda/4=30 cm$

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. $100\space ms^{-1}$


Correct Option: B
Explanation:

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

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

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


Correct Option: A
Explanation:

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

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

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. $256\ Hz$


Correct Option: B
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 $252\ Hz$

 

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. $3l _{2}-l _{1}$


Correct Option: A

Two pendulums of length $1.21m$ and $1.0m$ start vibrating. At some instant, the two are in the mean position in same phase. After how many vibrations of the longer pendulum, the two will be in phase?

  1. $10$

  2. $11$

  3. $20$

  4. $21$


Correct Option: B

A sound wave of wavelength $\lambda$ travels towards the right horizontally with a velocity $V$. It strikes and reflects from a vertical plane surface, traveling at a speed $v$ towards the left. The number of positive crests striking in a time interval of $3s$ on the wall is:

  1. $3(V+v)/ \lambda$

  2. $3(V-v)/ \lambda$

  3. $(V+v)/3\lambda$

  4. $(V-v)/3\lambda$


Correct Option: A

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


Correct Option: A
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.

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


Correct Option: B
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).

In Kundt's tube experiment wavelength in the metallic rod and air are 80 cm and 16 cm respectively. If the velocity of sound in air is $\displaystyle 300   ms^{-1}$ then the velocity of sound in rod will be

  1. $\displaystyle 80 ms^{-1}$

  2. $\displaystyle 3.75 ms^{-1}$

  3. $\displaystyle 240 ms^{-1}$

  4. $\displaystyle 1500 ms^{-1}$


Correct Option: D
Explanation:

Velocity of sound in air $V _{air}=300 ms^{-1}$, and $\lambda _{air}=16 cm=0.16 m$.

let us say velocity of sound in metal pipe $V _{metal} ms^{-1}$, and 

$\lambda _{metal}=80 cm=0.8 m$.

frequency remain unchanged when medium changes,

 $V _{metal}=\frac{\lambda _{metal}}{\lambda _{air}} V _{air} ms^{-1}=(0.8/0.16)*300 ms^{-1}=1500 ms^{-1}$.

Option "D" is correct.

If in an experiment for determination of velocity of sound by resonance tube method using a tuning fork of 512 Hz, first resonance was observed at 30.7 cm and second was obtained at 63.2 cm , then maximum possible error in velocity of sound is ( consider actual speed of sound in air is 332 m/s )

  1. $204$ $cm/sec$

  2. $280$ $cm/sec$

  3. $58$ $cm/sec$

  4. $80$ $cm/sec$


Correct Option: B

Which is(are) the factor(s) on which the frequency of sound emitted due to vibration in an air column depends?

  1. Length of air column

  2. Diameter of air column

  3. Both A and B

  4. None of these


Correct Option: C
Explanation:

In a closed air column, the frequency of mth mode of vibration is given by:

          $f _{m}=\dfrac{v(2m-1)}{4l}$      m=1,2,3......,
In a open air column, the frequency of mth mode of vibration is given by:
          $f _{m}=\dfrac{vm}{2l}$      m=1,2,3......
We can see that frequency depends upon the length of air column $l$ .  
Rayleigh stated that in an air column the antinodes are not formed exactly at the open end, but slightly outside. The distance of the antinode from the antinode is called end correction (e), therefore the length of the air column will change due to end correction hence frequency, this end correction depends upon the radius, r (or diameter) of air column and is equal to $0.6r $.      
Therefore frequency of vibration depends upon on both length and diameter of air column .

The speed of sound waves depends on temperature but speed of light waves does not. Why?

  1. Sound requires medium to travel and light doesn't.

  2. Frequency of sound is less than frequency of light.

  3. Wavelength of sound is larger than wavelength of light.

  4. Speed of sound is smaller than speed of light.


Correct Option: A

Which of the following can be used to determine the velocity of sound in solids, liquids as well as in gases :

  1. resonance tube

  2. kundt's tube

  3. sonometer.

  4. organ pipe


Correct Option: B
Explanation:

Kundt's tube is used to determine the velocity of sound in solids, liquids as well as in gases. In Kundt's tube the longitudinal waves are produced in air column and in rod. The nodes are detected by powder particles and the velocity of sound is determined. Also, the sound velocity in liquids is determined using Kundt's tube. The resonance tube, organ pipes are used to determine velocity of sound in air only. And sonometer determines velocity of sound in solids only.  

In an experiment to measure the speed of sound by a resonating air column, a tuning fork of frequency $500Hz$ is used. The length of the air column is varied by changing the level of water in the resonance tube. Two successive resonancers are heard at air columns of length $50.7cm$ and $83.9cm$. Which of the following statements is (are) true?

  1. The speed of sound determined from this experiment is $332m{s}^{-1}$

  2. The end correction in this experiment is $0.9cm$

  3. The wavelength of the sound wave is $66.4cm$

  4. The resonance at $50.6cm$ corresponds to the fundamental harmonic


Correct Option: A
Explanation:

$\begin{array}{l} Let\, \, { n _{ 1 } }\, \, harmonic\, \, is\, \, corresponding\, \, to\, \, 50.7\, cm\, \, and\, \, { n _{ 2 } }\, \, harmonic\, \, is\, \, corresponding\, \, 83.9\, cm. \ { { Sin } }ce\, \, both\, \, one\, \, con\sec  utive\, \, harmonics \ \therefore their\, \, difference=\frac { \lambda  }{ 2 }  \ \therefore \frac { \lambda  }{ 2 } =\left( { 83.9-50.7 } \right) cm \ \frac { \lambda  }{ 2 } =33.2\, cm \ \lambda =66.4\, cm \ \therefore \frac { \lambda  }{ 4 } =16.6\, cm \ length\, \, corresponding\, \, to\, \, fundamental\, \, e\, \, must\, \, be\, \, close\, to\, \, \frac { \lambda  }{ 4 } \, and\, \, 50.7\, cm, \ must\, \, be\, \, and\, \, odd\, \, multiple\, \, of\, \, this\, \, length\, \, 16.6\times 3=49.8\, cm. \ Therefore,\, 50.7\, \, is\, \, { 3^{ rd } }\, \, harmonic \ e+50.7=\frac { { 3\lambda  } }{ 4 }  \ e=49.8-50.7=-0.9\, cm \ speed\, \, of\, \, sound,v=f\lambda  \ \therefore v=500\times 66.4\, cm/\sec  =332\, m/s \end{array}$

Hence,
option $(A)$ is correct answer.

An open pipe of length 33 cm resonates with frequency of 1000 Hz. If the speed of sound s 330 $ms^{-1}$, then this frequency is 

  1. The fundamental frequency of the pipe'

  2. The first harmonic of the pipe

  3. The second harmonic of the pipe

  4. The forth harmonic of the pipe


Correct Option: C
Explanation:

Fundamental frequency (first harmonic frequency) of an open organ pipe is given by ,

                  $n _{1}=v/\lambda=v/2l$ ,
where $v=$speed of sound in air ($=330m/s , given$),
           $\lambda=$ wavelength ,
           $l=$ length of organ pipe ($=33cm=0.33m$  , given) ,
therefore ,  $n _{1}=330/(2\times0.33)=500Hz$ ,
as an open pipe produces even and odd harmonics , and given frequency is 1000Hz , it is 2 times of the first harmonic frequency , therefore it is second harmonic frequency of the pipe .

In a resonance tube the first resonance with a tuning fork occurs at 16 cm and second at 49 cm. If the velocity of sound is 330 m/s, the frequency of tuning fork is :

  1. 500 Hz

  2. 300 Hz

  3. 330 Hz

  4. 165 Hz


Correct Option: A
Explanation:

For closed pipe $l _1=\dfrac{v}{4n}$
$v=2n(l _2-l _1)$
$n=\dfrac{v}{2(l _2-l _1)}$$=\dfrac{330}{2\times (0.49-0.16)}=500 Hz$

A note has a frequency $128\ Hz$. The frequency of a note two octaves higher than it is

  1. $256\ Hz$

  2. $64\ Hz$

  3. $32\ Hz$

  4. $512\ Hz$


Correct Option: D
Explanation:

The sum of the intervals between adjacent notes of the major diatonic scale is an octave.
A series of notes arranged, such that their fundamental frequencies have definite ratios is called a musical scale.
In $1588$, Zarlino constructed a musical scale by introducing six notes between an octave. These eight notes constitute major diatonic scale. The first note or the note of the lowest frequency is called keynote and ratio of the frequencies of the two notes is called interval between them. It means two octaves higher means four times the given frequency.
$\therefore$ Required frequency $=4\times 128$
$= 512\ Hz$

Which of the following is not correct regarding the experiment to determine the velocity of sound in laboratory by resonance tube method?

  1. the resonance tube apparatus should be kept inclined

  2. the turning fork should be struck gently against a soft rubber pad

  3. the vibrating tuning fork should be held horizontally over the open end of the tube

  4. the prongs of vibrating tuning fork should not touch the tube


Correct Option: A
Explanation:

The length of the air column inside the resonance tube is determined by the amount of water kept in the tube. 
If a water filled tube is inclined, the length of resonance column would be different at different lines parallel to the tube, and thus a number of resonating lengths would be created.

In ,a resonance tube the first resonance occurs at 16 cm and the second resonance occurs at 49 cm. The end corrections will be :

  1. 0.3 cm

  2. 0.5 cm

  3. 0.8 cm

  4. 1.0 cm


Correct Option: B
Explanation:

Answer is B.

The end correction is given as follows.
End corrections $\displaystyle =\frac { { l } _{ 2 }-3{ l } _{ 1 } }{ 2 }$
$\displaystyle =\frac { 49cm-3\times 16cm }{ 2 } =0.5cm$.
Hence, the end corrections will be 0.5 cm.

In the experiment to determine the speed of sound using a resonance column,

  1. prongs of the tuning fork are kept in a vertical plane

  2. prongs of the tuning fork are kept in a horizontal plane

  3. in one of the two resonances observed, the length of the resonating air column is close to the wavelength of sound

    in air

  4. in one of the two resonances observed, the length of the resonating air column is close to half of the wavelength of

    sound in air


Correct Option: A
Explanation:

    In an experiment to determine the speed of sound using a resonance column , prongs of the tuning fork are kept in a vertical plane so that both the prongs can send sound waves inside the tube .

  In one resonance the length of resonating air column is , 
                        $l _{1}=\lambda/4$ ,
and in another resonance the length of resonating air column is , 
                        $l _{2}=3\lambda/4$ ,
only option B is correct .

In a resonance column experiment, the first resonance is obtained when the level of the water in tube is $20 cm$ from the open end. Resonance will also be obtained when the water level is at a distance of

  1. $40 cm$ from the open end.

  2. $60$ cm from the open end.

  3. $80$ cm from the open end.

  4. data insufficient


Correct Option: B
Explanation:

Frequency of first resonance in closed pipe    $\nu = \dfrac{v}{4L _1}$      ....(1)
where $L _1 = 20 \ cm$
Frequency of next resonance in closed pipe  $\nu = \dfrac{3v}{4L _2}$      ....(2)
Equating (1) and (2), we get
$\dfrac{v}{4L _1} = \dfrac{3v}{4L _2}$
$\implies$  $L _2 = 3L _1  = 3\times 20 = 60 \ cm$

A resonance air column of length 20 cm resonates with a tuning fork of frequency 250 Hz. The speed of sound in air is

  1. 300 m/s

  2. 200 m/s

  3. 150 m/s

  4. 75 m/s


Correct Option: B
Explanation:

Length of air column  $l = 20 \ cm = 0.2 \  m$
In resonance air column,
$\lambda = 4l  = 4\times 0.2 = 0.8 \ m$
Now, $\displaystyle \lambda =0.8m$ and $\displaystyle \nu=250Hz$
Hence, speed of sound in air  $\displaystyle v=\nu\lambda =250\times 0.8=200{ m }/{ s }$

Consider the following statements regarding the experiment to the determine the velocity of sound in laboratory by resonance tube method.
1. The first resonance is obtained for the length ${x} _{1}$ of the air column
2. The second resonance is obtained for the length ${x} _{2}$ of the air column.
If $n$ be the frequency of the tuning fork then which is the correct relation ($v$ represents the velocity of sound) ?

  1. $v=2n({x} _{1}+{x} _{2})$

  2. $v=2n({x} _{1}-{x} _{2})$

  3. $v=2n({x} _{2}-{x} _{1})$

  4. $v=2n({x} _{1}{x} _{2})$


Correct Option: C
Explanation:

Frequency in a resonant tube=$f=p\dfrac{v}{2l}$

where $p$ is for the pth harmonic.
Thus $n=\dfrac{v}{2x _1}$
and $n=2\dfrac{v}{2x _2}$
$\implies v=2n(x _2-x _1)$

A wire of density $\rho $ is stretched between the clamps at a distance $L$ apart while being subjected to an extension $\ell (<<L)$, Y is Young's modulus of the wire. The lowest resonant frequency of transverse vibration of the wire is approximately given by :

  1. $f= \dfrac{1}{2L}\sqrt{\dfrac{YL}{\ell\rho }}$

  2. $f= \dfrac{1}{2L}\sqrt{\dfrac{Y\rho L}{\ell^2 }}$

  3. $f= \dfrac{1}{2L}\sqrt{\dfrac{Y\ell}{L\rho }}$

  4. $f= \dfrac{1}{2L}\sqrt{\dfrac{L\rho}{Y\ell }}$


Correct Option: C
Explanation:
$Y=\dfrac{Stress}{Strain}=\dfrac{TL}{A\ell}$

$T= \dfrac{YA\ell}{L}$

mass per unit length

$f= \dfrac{1}{2L}\sqrt{T/m} $

$\ m\Rightarrow mass\:  per \: unit\:  length$

$f= \dfrac{1}{2L}\sqrt{\dfrac{Y\ell}{L\rho }}$

- Hide questions