Determining wavelength and speed of sound - class-IX

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

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

Using the Doppler effect formula for a source approaching a stationary observer: f' = f * (v / (v - vs)). Given vs = v/3, f' = 200 * (v / (v - v/3)) = 200 * (v / (2v/3)) = 200 * (3/2) = 300 Hz.

The intensity of a sound wave is given by I = 2 * pi^2 * f^2 * A^2 * rho * v. Rearranging for amplitude A: A = sqrt(I / (2 * pi^2 * f^2 * rho * v)). Plugging in values: I = 2e-6, f = 1000, rho = 1.2, v = 330. A = sqrt(2e-6 / (2 * 9.87 * 1e6 * 1.2 * 330)) = 1.6e-8 m.

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.

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 is a classic laboratory apparatus used to determine the speed of sound in air by measuring the lengths of air columns that resonate with a tuning fork of known frequency.

The standard audible frequency range for a healthy human ear is generally accepted to be from 20 Hz to 20,000 Hz.

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

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

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

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

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 resonance tube, the resonance lengths are l1 = lambda/4, l2 = 3lambda/4, and l3 = 5lambda/4. The difference between consecutive resonances is lambda/2. Thus, l3 = l2 + (l2 - l1) = 2*l2 - l1.

The relative velocity of the wave crests with respect to the wall is (V + v). The distance covered in time t is (V + v) * t. The number of crests is distance / wavelength = (V + v) * t / lambda. For t = 3s, this is 3(V + v) / lambda.

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).

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

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

Sound is a mechanical wave requiring a medium, so its speed depends on the medium's properties like temperature and density. Light is an electromagnetic wave that can travel through a vacuum, and its speed is a fundamental constant independent of the medium's temperature.

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.  

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

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

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$

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 $=4\times 128$
$= 512\ Hz$

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.

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 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 .

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
$\dfrac{v}{4L _1} = \dfrac{3v}{4L _2}$
$\implies$  $L _2 = 3L _1  = 3\times 20 = 60 \ cm$

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