Waves

Fundamental frequency of open pipe (at both ends) $f=\frac{V}{2 L} \ldots$ (i)

Now immersed in water open pipe behaves as closed pipe.

$\begin{aligned} & f^{\prime}=\frac{v}{4\left(\frac{L}{2}\right)}=\frac{v}{2 L} \quad\text{..... (ii)}\\ & f=f^{\prime} \end{aligned}$

To find the frequency of the wave, we need to start by analyzing the given displacement equation of the wave:

$y = C \sin \left( \frac{2 \pi}{\lambda} (at - x) \right)$

Where:

• $y$ is the displacement of the wave
• $C$ is the amplitude
• $\frac{2 \pi}{\lambda}$ is the wave number (denoting how many wavelengths fit into a unit length)
• $\lambda$ is the wavelength
• $a$ is some constant (likely representing the speed of the wave)
• $t$ is time
• $x$ is the distance

By comparing with the standard form of a travelling wave, we have:

$y = C \sin \left( k (at - x) \right)$

Where $k$ is the wave number:

$k = \frac{2 \pi}{\lambda}$

From the standard wave equation, the argument of the sine function is usually written as:

$k (at - x)$

This implies that $ka$ in the wave equation represents the angular frequency $\omega$ of the wave:

$\omega = k a$

Substituting $k = \frac{2 \pi}{\lambda}$ into $\omega$:

$\omega = \left( \frac{2 \pi}{\lambda} \right) a = \frac{2 \pi a}{\lambda}$

Angular frequency $\omega$ is related to the frequency $f$ by:

$\omega = 2 \pi f$

So:

$2 \pi f = \frac{2 \pi a}{\lambda}$

Solving for the frequency $f$:

$f = \frac{a}{\lambda}$

Therefore, the correct answer is:

Option D: $\frac{a}{\lambda}$

$\begin{aligned} & \mathrm{n}_{\text {oop }}=(2 \mathrm{M}+1)^{\text {th }} \text { Har. }=(2 \times 4+1) \times \frac{\mathrm{V}}{4 \ell_{\mathrm{c}}}=\frac{9 \mathrm{~V}}{4 \ell_{\mathrm{c}}} \\ & \mathrm{n}_{\text {oop }}=(\mathrm{M}+1)^{\text {th }} \text { Har. }=(3+1) \frac{\mathrm{V}}{2 \ell_0}=\frac{4 \mathrm{~V}}{2 \ell_0} \\ & \text { now } \frac{9 \mathrm{~V}}{4 \ell_{\mathrm{c}}}=\frac{4 \mathrm{~V}}{2 \ell_0} \\ & \frac{\ell_{\mathrm{c}}}{\ell_0}=\frac{18}{16}=\frac{9}{8} \end{aligned} $

The fundamental frequency of an open pipe, n_oop, is given by V / (2L), where V is the speed of sound and L is the length of the pipe.

For a closed pipe of the same length, the fundamental frequency, n_cop, is V / (4L) since only half as many wavelengths fit into the same length due to the closed end.

Therefore, the ratio of frequencies of an open pipe to a closed pipe is:

$\frac{n_{\text{oop}}}{n_{\text{cop}}} = \frac{V / (2L)}{V / (4L)} = \frac{4L}{2L} = 2:1$

This shows that the fundamental frequency of an open pipe is twice that of a closed pipe of the same length.

${f_0} = {v \over {2L}} = {{\sqrt {{{\gamma RT} \over M}} } \over {2L}}$

${{{f_{{0_1}}}} \over {{f_{{0_2}}}}} = \sqrt {{{{T_1}} \over {{T_2}}}} = \sqrt {{{300} \over {363}}} = {1 \over {1.1}}$

${f_{{0_2}}} = 1\,.\,1{f_{{0_1}}} = 1\,.\,1(400)$

$ = 440$ Hz

We know, velocity of transverse wave

$v = \sqrt {{T \over \mu }} $

$\therefore$ ${v_i} = \sqrt {{T \over \mu }} $ and ${v_f} = \sqrt {{{2T} \over \mu }} $

$\therefore$ ${{{v_i}} \over {{v_f}}} = {1 \over {\sqrt 2 }}$