Acoustics and Sound Waves

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

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 .

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  $\displaystyle v=\nu\lambda =250\times 0.8=200{ m }/{ s }$

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

In vibrating motion increase in amplitude results in increase in loudness.

Statement C is true, consider the case of a guitar string mounted to the sound box. The fact that the sound box is greater than the surface area of the string means more surrounding particles will be forced into vibration causes an increase in amplitude and loudness.

Damped vibrations in which an oscillating system has the effect of reducing, restricting or preventing its oscillations.

Damping is caused by opposing force, which decreases the frequency.

Ans: C

The time taken by vibrating body to complete one vibration is called time period. so our given statement is false.

Earthquake is a movement of tremor of the earth's crust. It originated naturally and below the surface. An earthquake produces infrasound before the mainshock wave begins. Infrasound is low-frequency sound less than 20hz. It can travel hundreds of km through the earth's surface. So by detecting this sound, we can predict an earthquake.