Tag: resonance and sonometer

Questions Related to resonance and sonometer

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

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

  1. Increases

  2. Decreases

  3. Remains the same

  4. None of these

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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 

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

  1. transverse vibration.

  2. longitudinal vibrations.

  3. both.

  4. none of these

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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 

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

  1. $450 Hz$

  2. $300 Hz$

  3. $400 Hz$

  4. $425 Hz$

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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)

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

  1. $4.4 \times 10 ^ { - 8 } m$

  2. $1.6 \times 10 ^ { - 8 } m$

  3. $2.4 \times 10 ^ { - 6 } m$

  4. $1.8 \times 10 ^ { - 6 } m$

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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?

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

  1. 25Hz

  2. 10Hz

  3. 40Hz

  4. 250Hz

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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

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

  1. $4\times 10^{-4}$ $m$

  2. $8\times 10^{-4}$ $m$

  3. $4\times 10^{-3}$ $m$

  4. $8\times 10^{-3}$ $m$

Show answer
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.

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

  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

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Correct Option: D

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

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

  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

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Correct Option: B

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

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

  1. $65$

  2. $75$

  3. $80$

  4. $90$

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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 ..........

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

  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 }) } $

Show answer
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 $=\dfrac{1}{2(n _1-n _2)}$