Tag: resonance and sonometer

Questions Related to resonance and sonometer

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

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

  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

Show answer
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

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

  1. $40 cm$ from the open end.

  2. $60$ cm from the open end.

  3. $80$ cm from the open end.

  4. data insufficient

Show answer
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

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

  1. 300 m/s

  2. 200 m/s

  3. 150 m/s

  4. 75 m/s

Show answer
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) ?

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

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

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

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

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

Show answer
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 :

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

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

Show answer
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} $

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

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