Tag: resonance tube

Questions Related to resonance tube

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:

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

  1. $20\ cm$

  2. $30\ cm$

  3. $40\ cm$

  4. None of the above

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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

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

  1. $660\space ms^{-1}$

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

  3. $230\space ms^{-1}$

  4. $100\space ms^{-1}$

Show answer
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?

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

  1. $87.5$m

  2. $700$m

  3. $1400$m

  4. $1.75$m

Show answer
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 : 

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

  1. $250\ Hz$

  2. $252\ Hz$

  3. $260\ Hz$

  4. $256\ Hz$

Show answer
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

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

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

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

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

  4. $3l _{2}-l _{1}$

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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?

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

  1. $10$

  2. $11$

  3. $20$

  4. $21$

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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:

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

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

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

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

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

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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

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

  1. 8

  2. 800

  3. 7

  4. 6

Show answer
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 :

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

  1. amplitude

  2. frequency

  3. wavelength

  4. particle velocity

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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

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

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

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

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

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

Show answer
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.