(4)

The fundamental frequency $\left(f\right)$ is related to the length of the pipe $\left(L\right)$ and speed of sound in air $\left(v\right)$ as

          $f=\frac{v}{\lambda }$

For a pipe open at both ends, the fundamental frequency corresponding to a wavelength $\lambda$ that is twice the length of the pipe is

        $f=\frac{v}{2L}$                                                           ...(i)

When the pipe is dipped vertically in water so that half of its length is submerged, the new length of the air column becomes

        $L\text{'}=\frac{L}{2}$                                                           ...(ii)

For the new length of the air column, the wavelength $\lambda \text{'}$ is given by

          $\lambda \text{'}=4L\text{'}$

         $\lambda \text{'}=4\times \frac{L}{2}$                                  $\text{(Using (ii))}$

         $\lambda \text{'}=2L$

The new fundamental frequency,

          $f\text{'}=\frac{v}{\lambda \text{'}}=\frac{v}{2L}$                                                     ...(iii)

Comparing eq. (i) and (iii), we get

        $f\text{'}=f$

(3)

$4^{\text{th}}$ overtone of closed organ pipe $=9{\upsilon }_0=\frac{9\times \nu }{4L}$              ...(i)

3rd overtone of open organ pipe $=4{\upsilon }_0=\frac{4\times \nu }{2L\text{'}}$                  ...(ii)

As per question, $9{\upsilon }_0=4{\upsilon }_0$

From equations (i) and (ii), $\frac{9\times \nu }{4L}=\frac{4\times \nu }{2L\text{'}}$

$\Rightarrow \frac{L}{L\text{'}}=\frac{9}{8}$

(4)

Fundamental frequency for open organ pipe,

$f_o=\frac{v}{2L_o}$                                             ...(i)

Fundamental frequency for closed organ pipe,

$f_c=\frac{v}{4L_c}$                                            ...(ii)

$\frac{f_o}{f_c}=\frac{v}{2L_o}\times \frac{4L_c}{v}\Rightarrow \frac{f_o}{f_c}=\frac{2}{1}$         $(\because L_o=L_c\text{ (given)})$

(3)

Frequency $\left(\upsilon \right)$ = 800 Hz

As the pipe is closed at one end, so

$l_3-l_2=l_2-l_1=\frac{\lambda }{2}=21.5$$\text{cm}$

$\therefore$   $\lambda =43.0$ $\text{cm}$

As $\upsilon =\frac{v}{\lambda }\Rightarrow v=\upsilon \lambda$

$\therefore$   $v=\frac{800\times 43}{100}=344$$\text{m}$${\text{s}}^{-1}$

(2)

The velocity of sound in air at $27°$C is $v=2\left(\upsilon \right)[L_2-L_1];$

where $\upsilon =$frequency of tuning fork and $L_1,L_2$ are the successive column lengths.

$\therefore$ $v=2\times 320[73-20]\times {10}^{-2}$

          $=339.2$$\text{m}$ ${\mathrm{s}}^{-1}\approx 339$$\text{m}$ ${\mathrm{s}}^{-1}$

(1)

For closed organ pipe, third harmonic is $\frac{3v}{4l}$

For open organ pipe, fundamental frequency is $\frac{v}{2l\text{'}}$

Given, third harmonic for closed organ pipe = fundamental frequency for open organ pipe.

$\therefore$   $\frac{3v}{4l}=\frac{v}{2l\text{'}}\Rightarrow l\text{'}=\frac{4l}{3\times 2}=\frac{2l}{3};$

where $l$ and $l\text{'}$ are the lengths of closed and open organ pipes respectively.

$\therefore$    $l\text{'}=\frac{2\times 20}{3}=13.33$ $\text{cm}$

(1)

Nearest harmonics of an organ pipe closed at one end differ by twice of its fundamental frequency.

$\therefore$   $260-220=2\upsilon \Rightarrow \upsilon =20$ $\text{Hz}$

(2)

Second overtone of an open organ pipe

          (Third harmonic) $=3\times {\upsilon }_0\text{'}=3\times \frac{v}{2L\text{'}}$

First overtone of a closed organ pipe

          (Third harmonic) $=3\times {\upsilon }_0=3\times \frac{v}{4L}$

According to question,

          $3{\upsilon }_0\text{'}=3{\upsilon }_0\Rightarrow 3\times \frac{v}{2L\text{'}}=3\times \frac{v}{4L}$

$\therefore$    $L\text{'}=2L$

(1)

From Figure,

First harmonic is obtained at

           $l=\frac{\lambda }{4}=50$ $\text{cm}$

Third harmonic is obtained for resonance,

         $l\text{'}=\frac{3\lambda }{4}=3\times 50=150$$\text{cm}$

(2)

For a string fixed at both ends, the resonant frequencies are

${\upsilon }_n=\frac{nv}{2L} \text{where }n=1,2,3,\dots$

The difference between two consecutive resonant frequencies is

$\Delta {\upsilon }_n={\upsilon }_{n+1}-{\upsilon }_n=\frac{(n+1)v}{2L}-\frac{nv}{2L}=\frac{v}{2L}$

which is also the lowest resonant frequency ($n=1$).
Thus, the lowest resonant frequency for the given string

= 420 Hz - 315 Hz = 105 Hz