Transverse Stationary Waves and Sonometer Questions in English

(D) The frequency of the sonometer wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $L$ is the length of the loop,$T$ is the tension,and $\mu$ is the linear mass density.
Since the frequency $f$ remains constant,we have $\frac{1}{L_1} \sqrt{T_1} = \frac{1}{L_2} \sqrt{T_2}$.
Given $\frac{L_1}{L_2} = 2.2$,we have $\sqrt{\frac{T_1}{T_2}} = 2.2$,which implies $T_1 = (2.2)^2 T_2 = 4.84 T_2$.
When the cylinder is in air,$T_1 = Mg$. When immersed in water,$T_2 = Mg - F_B$,where $F_B$ is the buoyant force.
$F_B = V_{cyl} \rho_w g = \frac{M}{\rho_{cyl}} \rho_w g = \frac{Mg}{S.G.}$,where $S.G.$ is the specific gravity.
Thus,$Mg = 4.84 (Mg - \frac{Mg}{S.G.})$.
Dividing by $Mg$,we get $1 = 4.84 (1 - \frac{1}{S.G.})$.
$1 = 4.84 - \frac{4.84}{S.G.}$,which gives $\frac{4.84}{S.G.} = 3.84$.
$S.G. = \frac{4.84}{3.84} \approx 1.26$.

(C) In a stationary wave,the points where the amplitude of vibration is zero are called nodes,and the points where the amplitude is maximum are called antinodes.
Paper pieces placed at antinodes will fall off due to maximum vibration,while a paper piece placed at a node will remain on the wire because the amplitude of vibration at a node is zero.
Since the middle piece of paper does not fall,it must be located at a node.
The distance between the two supports $Q$ and $R$ is $L = 4x$.
Given that the middle piece is at a node,the wire vibrates in two loops (each of length $2x$).
The length of one loop is $\frac{\lambda}{2}$.
Therefore,$2x = \frac{\lambda}{2}$,which gives $\lambda = 4x$.
Thus,the wavelength of the vibration is $4x$.

(B) The fundamental frequency of the original wire of length $l$ is $n_0 = \frac{v}{2l}$,where $v$ is the wave speed.
When a bridge is introduced at a distance $\Delta l$ from the center,the wire is divided into two segments of lengths $l_1 = \frac{l}{2} + \Delta l$ and $l_2 = \frac{l}{2} - \Delta l$.
The fundamental frequencies of these two segments are:
$n_1 = \frac{v}{2l_1} = \frac{v}{2(\frac{l}{2} + \Delta l)} = \frac{v}{l + 2\Delta l}$
$n_2 = \frac{v}{2l_2} = \frac{v}{2(\frac{l}{2} - \Delta l)} = \frac{v}{l - 2\Delta l}$
The beat frequency is $|n_2 - n_1| = \frac{v}{l - 2\Delta l} - \frac{v}{l + 2\Delta l}$.
Using the binomial approximation $(1 \pm x)^{-1} \approx 1 \mp x$ for $x \ll 1$:
$|n_2 - n_1| = \frac{v}{l} \left( (1 - \frac{2\Delta l}{l})^{-1} - (1 + \frac{2\Delta l}{l})^{-1} \right)$
$\approx \frac{v}{l} \left( (1 + \frac{2\Delta l}{l}) - (1 - \frac{2\Delta l}{l}) \right)$
$= \frac{v}{l} \left( \frac{4\Delta l}{l} \right) = \frac{4v}{l} \cdot \frac{\Delta l}{l}$.
Since $n_0 = \frac{v}{2l}$,we have $\frac{v}{l} = 2n_0$.
Substituting this,the beat frequency is $4(2n_0) \frac{\Delta l}{l} = 8 n_0 \frac{\Delta l}{l}$.

(C) The fundamental frequency of a sonometer wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$.
Case $1$: Tension is increased by $44\%$. The new tension $T' = T + 0.44T = 1.44T$.
The new frequency $n' = n + 6$.
$n + 6 = \frac{1}{2l} \sqrt{\frac{1.44T}{m}} = 1.2 \left( \frac{1}{2l} \sqrt{\frac{T}{m}} \right) = 1.2n$.
$0.2n = 6 \Rightarrow n = 30 \ Hz$.
Case $2$: Length is increased by $20\%$. The new length $l' = l + 0.2l = 1.2l$.
The new frequency $n'' = \frac{1}{2l'} \sqrt{\frac{T}{m}} = \frac{1}{2(1.2l)} \sqrt{\frac{T}{m}} = \frac{n}{1.2} = \frac{30}{1.2} = 25 \ Hz$.
The change in frequency is $|n - n''| = |30 - 25| = 5 \ Hz$.

(C) The frequency $f$ of a standing wave in a wire under tension $T$ is given by $f = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$,where $n$ is the number of antinodes,$L$ is the length of the wire,and $\mu$ is the linear mass density.
The tension $T$ in the wire is provided by the spring,so $T = kx$,where $k$ is the spring constant and $x$ is the stretch.
Thus,$f = \frac{n}{2L} \sqrt{\frac{kx}{\mu}}$.
For the first case,$n_1 = 3$ and $x_1 = 4.0\, cm$. So,$f = \frac{3}{2L} \sqrt{\frac{k(4.0)}{\mu}}$.
For the second case,$n_2 = 2$ and we need to find $x_2$. Since the frequency $f$ is the same,$f = \frac{2}{2L} \sqrt{\frac{k(x_2)}{\mu}}$.
Equating the two expressions for $f$: $\frac{3}{2L} \sqrt{\frac{k(4.0)}{\mu}} = \frac{2}{2L} \sqrt{\frac{k(x_2)}{\mu}}$.
Simplifying,we get $3 \sqrt{4.0} = 2 \sqrt{x_2}$.
$3 \times 2 = 2 \sqrt{x_2} \implies 6 = 2 \sqrt{x_2} \implies \sqrt{x_2} = 3$.
Squaring both sides,$x_2 = 9\, cm$.

(D) The fundamental frequency of a stretched wire is given by $\mu = \frac{1}{2L} \sqrt{\frac{F}{\lambda}}$,where $\lambda$ is the mass per unit length.
Given that the total mass is $M$ and length is $L$,the mass per unit length $\lambda = \frac{M}{L}$.
Substituting this into the frequency formula:
$\mu = \frac{1}{2L} \sqrt{\frac{F}{M/L}}$
$\mu = \frac{1}{2L} \sqrt{\frac{FL}{M}}$
$\mu = \frac{1}{2} \sqrt{\frac{FL}{M \cdot L^2}}$
$\mu = \frac{1}{2} \sqrt{\frac{F}{ML}}$
Resonance occurs when the impressed frequency matches the fundamental frequency of the wire,leading to maximum vibrational energy.

(D) The fundamental frequency of a vibrating string is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Linear mass density $\mu = \text{mass per unit length} = \rho \times \text{Area of cross-section} = \rho \pi r^2$,where $\rho$ is the density of the material.
Given: $L_1 = L$,$r_1 = 2r$,$L_2 = 2L$,$r_2 = r$,and $T_1 = T_2 = T$.
For the first string: $f_1 = \frac{1}{2L_1} \sqrt{\frac{T}{\rho \pi r_1^2}} = \frac{1}{2L} \sqrt{\frac{T}{\rho \pi (2r)^2}} = \frac{1}{4L} \sqrt{\frac{T}{\rho \pi r^2}}$.
For the second string: $f_2 = \frac{1}{2L_2} \sqrt{\frac{T}{\rho \pi r_2^2}} = \frac{1}{2(2L)} \sqrt{\frac{T}{\rho \pi r^2}} = \frac{1}{4L} \sqrt{\frac{T}{\rho \pi r^2}}$.
Taking the ratio: $\frac{f_1}{f_2} = \frac{\frac{1}{4L} \sqrt{\frac{T}{\rho \pi r^2}}}{\frac{1}{4L} \sqrt{\frac{T}{\rho \pi r^2}}} = 1$.

(A) The cord vibrates in its simplest standing-wave mode (fundamental mode),where the length of the cord $d$ corresponds to half a wavelength,i.e.,$d = \frac{\lambda}{2}$,or $\lambda = 2d$.
The frequency of the standing wave $f$ is equal to the frequency of the $AC$ current.
The wave speed $v$ on a stretched string is given by $v = \sqrt{\frac{T}{\mu}}$.
Using the wave equation $v = f\lambda$,we substitute $\lambda = 2d$ to get $v = f(2d) = 2fd$.
Equating the two expressions for $v$: $\sqrt{\frac{T}{\mu}} = 2fd$.
Squaring both sides: $\frac{T}{\mu} = 4f^2 d^2$.
Therefore,the relationship is $T = 4\mu f^2 d^2$.

(B) The fundamental frequency of a stretched wire is given by $f = \frac{V}{2L}$,which implies $f \propto \frac{1}{L}$.
Given the ratio of frequencies $f_1 : f_2 : f_3 = 1 : 2 : 3$,the lengths of the segments must be in the ratio of the reciprocals of the frequencies:
$L_1 : L_2 : L_3 = \frac{1}{1} : \frac{1}{2} : \frac{1}{3} = 6 : 3 : 2$.
Let the lengths be $L_1 = 6x$,$L_2 = 3x$,and $L_3 = 2x$.
The total length is $L_1 + L_2 + L_3 = 110\; cm$.
$6x + 3x + 2x = 110 \Rightarrow 11x = 110 \Rightarrow x = 10\; cm$.
Thus,the lengths are $L_1 = 60\; cm$,$L_2 = 30\; cm$,and $L_3 = 20\; cm$.
The first bridge is placed at $L_1 = 60\; cm$ from $A$.
The second bridge is placed at $L_1 + L_2 = 60 + 30 = 90\; cm$ from $A$.

(A) The frequency of a sonometer wire is given by $n' \propto \frac{1}{l}$.
Let the frequency of the tuning fork be $n$.
For length $l_1 = 50 \, cm$,the frequency of the wire is $n_1 = \frac{k}{50}$. Since it produces $4$ beats,$n_1 = n \pm 4$.
For length $l_2 = 49 \, cm$,the frequency of the wire is $n_2 = \frac{k}{49}$. Since it produces $4$ beats,$n_2 = n \pm 4$.
Since $l_2 < l_1$,$n_2 > n_1$. Thus,$n_1 = n - 4$ and $n_2 = n + 4$.
So,$\frac{n_1}{n_2} = \frac{49}{50} = \frac{n-4}{n+4}$.
$50(n-4) = 49(n+4)$.
$50n - 200 = 49n + 196$.
$n = 396 \, Hz$.

(D) For a string fixed at both ends,the frequency of the $n^{th}$ harmonic is given by $f_n = \frac{n v}{2L}$,where $n = 1, 2, 3, \dots$ and $v = \sqrt{\frac{T}{\mu}}$.
The $3^{rd}$ overtone corresponds to the $4^{th}$ harmonic,so $n = 4$.
Given: Length $L = 1 \, m$,Tension $T = 200 \, N$,Linear mass density $\mu = 5 \, g/m = 5 \times 10^{-3} \, kg/m$.
First,calculate the wave speed $v = \sqrt{\frac{200}{5 \times 10^{-3}}} = \sqrt{40000} = 200 \, m/s$.
Now,calculate the frequency $f_4 = \frac{4 \times 200}{2 \times 1} = 2 \times 200 = 400 \, Hz$.

(B) Given: Length $L = 1.5 \, m$,Density $\rho = 7.7 \times 10^3 \, kg/m^3$,Young's modulus $Y = 2 \times 10^{11} \, N/m^2$,Strain $\epsilon = \frac{\Delta L}{L} = 1 \% = 0.01$.
Young's modulus is defined as $Y = \frac{F/A}{\Delta L/L}$,so the tension $F = Y \cdot A \cdot \epsilon$.
The fundamental frequency $n$ of a stretched string is given by $n = \frac{1}{2L} \sqrt{\frac{F}{\mu}}$,where $\mu = A \cdot \rho$ is the mass per unit length.
Substituting $F$: $n = \frac{1}{2L} \sqrt{\frac{Y \cdot A \cdot \epsilon}{A \cdot \rho}} = \frac{1}{2L} \sqrt{\frac{Y \cdot \epsilon}{\rho}}$.
Substituting the values: $n = \frac{1}{2 \times 1.5} \sqrt{\frac{2 \times 10^{11} \times 0.01}{7.7 \times 10^3}}$.
$n = \frac{1}{3} \sqrt{\frac{2 \times 10^9}{7.7 \times 10^3}} = \frac{1}{3} \sqrt{0.2597 \times 10^6} = \frac{1}{3} \times 509.6 \approx 170 \, Hz$.

(A) For the second harmonic in a string of length $l$ fixed at both ends,the standing wave pattern consists of two loops with a node at the center $(l/2)$.
To excite a specific harmonic,we pluck the string at an antinode and touch it at a node.
For the second harmonic $(n=2)$,the wavelength is $\lambda_2 = 2l/n = l$.
The nodes are located at $x = 0, l/2, l$.
The antinodes are located at $x = l/4$ and $x = 3l/4$.
To generate the second harmonic,we must pluck the string at an antinode (e.g.,$l/4$) and touch it at a node (e.g.,$l/2$) to suppress the fundamental frequency and allow the second harmonic to persist.

(D) The fundamental frequency $n$ of a sonometer wire is given by the formula: $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $\mu = \pi r^2 \rho$ is the mass per unit length.
Thus,$n = \frac{1}{2l} \sqrt{\frac{T}{\pi r^2 \rho}} = \frac{1}{2lr} \sqrt{\frac{T}{\pi \rho}}$.
From this,we see that $n \propto \frac{\sqrt{T}}{lr}$.
Let the initial values be $l_1 = l$,$T_1 = T$,and $r_1 = r$. The new values are $l_2 = 3l$,$T_2 = 3T$,and $r_2 = 3r$.
The ratio of the frequencies is given by:
$\frac{n_2}{n_1} = \sqrt{\frac{T_2}{T_1}} \times \frac{l_1}{l_2} \times \frac{r_1}{r_2}$
$\frac{n_2}{n} = \sqrt{\frac{3T}{T}} \times \frac{l}{3l} \times \frac{r}{3r}$
$\frac{n_2}{n} = \sqrt{3} \times \frac{1}{3} \times \frac{1}{3} = \frac{\sqrt{3}}{9} = \frac{1}{3\sqrt{3}}$
Therefore,$n_2 = \frac{n}{3\sqrt{3}}$.

(B) The resonant frequencies of a string fixed at both ends are given by $f_n = n \cdot f_0$,where $f_0$ is the fundamental frequency and $n$ is an integer $(n = 1, 2, 3, \dots)$.
Given two consecutive resonant frequencies $f_n = 315\,Hz$ and $f_{n+1} = 420\,Hz$.
The difference between two consecutive resonant frequencies is the fundamental frequency $f_0$.
$f_0 = f_{n+1} - f_n = 420\,Hz - 315\,Hz = 105\,Hz$.
Thus,the lowest resonant frequency (fundamental frequency) is $105\,Hz$.

(A) The fundamental frequency of a string fixed at both ends is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$,where $l$ is the length,$T$ is the tension,and $m$ is the mass per unit length.
For option $A$: $n' = \frac{1}{2(2l)} \sqrt{\frac{4T}{m}} = \frac{1}{4l} \cdot 2 \sqrt{\frac{T}{m}} = n$ (No change).
For option $B$: $n' = \frac{1}{2(l/2)} \sqrt{\frac{2T}{m}} = \sqrt{2} n$ (Changes).
For option $C$: $n' = \frac{1}{2(l/2)} \sqrt{\frac{T/2}{m}} = \frac{1}{\sqrt{2}} n$ (Changes).
For option $D$: $n' = \frac{1}{2(2l)} \sqrt{\frac{2T}{m}} = \frac{1}{\sqrt{2}} n$ (Changes).
Thus,the adjustment in option $A$ does not affect the fundamental frequency.

(B) The fundamental frequency $n$ of a sonometer wire is given by the formula $n = \frac{1}{2 \ell} \sqrt{\frac{T}{m}}$,where $\ell$ is the length,$T$ is the tension,and $m$ is the mass per unit length.
Since $\ell$ and $m$ are constant,we have $n \propto \sqrt{T}$.
Let the initial frequency be $n$. When the tension $T$ is increased by $44\%$,the new tension $T' = T + 0.44T = 1.44T$.
The new frequency $n'$ is given as $n + 6$.
Using the proportionality $n \propto \sqrt{T}$,we get $\frac{n'}{n} = \sqrt{\frac{T'}{T}}$.
Substituting the values: $\frac{n+6}{n} = \sqrt{\frac{1.44T}{T}} = \sqrt{1.44} = 1.2$.
Thus,$n + 6 = 1.2n$.
$0.2n = 6$.
$n = \frac{6}{0.2} = 30\, Hz$.

(C) The frequency of a sonometer wire is given by $n = \frac{1}{2 \ell} \sqrt{\frac{T}{\mu}}$,where $T = Mg$ is the tension in the wire and $\mu$ is the linear mass density.
Thus,$n = \frac{1}{2 \ell} \sqrt{\frac{Mg}{\mu}}$.
For the wire to be in resonance with the same tuning fork,the frequency $n$ must remain constant.
Since $\ell$ and $\mu$ are constant,we have $n \propto \sqrt{Mg}$.
Therefore,$\sqrt{M_1 g_1} = \sqrt{M_2 g_2}$.
Given $M_1 = 1 \, kg$,$g_1 = g$,and $g_2 = g/6$.
Substituting these values: $\sqrt{1 \times g} = \sqrt{M_2 \times \frac{g}{6}}$.
Squaring both sides: $g = M_2 \times \frac{g}{6}$.
Solving for $M_2$,we get $M_2 = 6 \, kg$.

(B) For a stretched wire under constant tension and linear mass density, the frequency $n$ is inversely proportional to the length $l$, i.e., $n \propto 1/l$ or $nl = \text{constant}$.
Given the total length $L = l_1 + l_2 + l_3 = 110 \, cm$.
The frequencies are in the ratio $n_1 : n_2 : n_3 = 1 : 2 : 3$.
Since $n_1 l_1 = n_2 l_2 = n_3 l_3 = k$, we have $l_1 = k/n_1, l_2 = k/n_2, l_3 = k/n_3$.
Thus, $l_1 : l_2 : l_3 = 1/1 : 1/2 : 1/3 = 6 : 3 : 2$.
Let the lengths be $6x, 3x,$ and $2x$.
Then $6x + 3x + 2x = 110 \, cm \Rightarrow 11x = 110 \, cm \Rightarrow x = 10 \, cm$.
Therefore, the lengths are $l_1 = 6(10) = 60 \, cm$, $l_2 = 3(10) = 30 \, cm$, and $l_3 = 2(10) = 20 \, cm$.

(D) The tension $T$ in the wire is equal to the weight of the hanging mass,$T = Mg = 9 \times 9.8 = 88.2 \, N$.
The linear mass density $\mu$ is $\mu = \frac{m}{l} = \frac{12 \times 10^{-3} \, kg}{1.5 \, m} = 8 \times 10^{-3} \, kg/m$.
The velocity of the transverse wave is $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{88.2}{8 \times 10^{-3}}} = \sqrt{11025} = 105 \, m/s$.
For vibration in $p=3$ loops,the frequency $f$ is given by $f = \frac{p v}{2l}$.
Substituting the values,$f = \frac{3 \times 105}{2 \times 1.5} = \frac{315}{3} = 105 \, Hz$.
Since $105 \, Hz$ is not among the options,the correct choice is $D$.

(C) The Young's modulus $Y$ is given by $Y = \frac{T/A}{\Delta \ell / \ell}$,where $T$ is tension,$A$ is cross-sectional area,$\Delta \ell$ is extension,and $\ell$ is length.
$T = \frac{Y A \Delta \ell}{\ell}$.
The velocity of transverse waves in the wire is $v = \sqrt{\frac{T}{\mu}}$,where $\mu = A \rho$ is the mass per unit length and $\rho$ is the density.
$v = \sqrt{\frac{Y A \Delta \ell}{\ell A \rho}} = \sqrt{\frac{Y \Delta \ell}{\ell \rho}}$.
Substituting the given values: $v = \sqrt{\frac{9 \times 10^{10} \times 4.9 \times 10^{-4}}{1 \times 9 \times 10^3}} = \sqrt{\frac{44.1 \times 10^6}{9 \times 10^3}} = \sqrt{4.9 \times 10^3} = \sqrt{4900} = 70 \,m/s$.
The lowest frequency (fundamental frequency) is $f = \frac{v}{2\ell} = \frac{70}{2 \times 1} = 35 \,Hz$.

(A) The given equation is $y = 0.06 \sin \left( \frac{2\pi}{3}x \right) \cos (120\pi t)$.
Comparing this with the standard wave equation,we have angular frequency $\omega = 120\pi \, rad/s$ and wave number $k = \frac{2\pi}{3} \, m^{-1}$.
The wave speed $v$ is given by $v = \frac{\omega}{k} = \frac{120\pi}{2\pi/3} = 180 \, m/s$.
The linear mass density $\mu$ is $\mu = \frac{m}{L} = \frac{3.0 \times 10^{-2} \, kg}{1.5 \, m} = 2.0 \times 10^{-2} \, kg/m$.
The tension $T$ in the string is given by $T = v^2 \mu$.
Substituting the values,$T = (180)^2 \times (2.0 \times 10^{-2}) = 32400 \times 0.02 = 648 \, N$.

(C) The fundamental frequency of a sonometer wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension (weight) and $\mu$ is the mass per unit length.
Since $f \propto \sqrt{T}$,we have $\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1}}$.
An octave means the frequency is doubled,so $f_2 = 2f_1$.
Therefore,$\frac{2f_1}{f_1} = \sqrt{\frac{T_2}{T_1}}$,which implies $2 = \sqrt{\frac{T_2}{T_1}}$.
Squaring both sides,we get $4 = \frac{T_2}{T_1}$.
Given $T_1 = 4\,kg-wt$,we find $T_2 = 4 \times 4 = 16\,kg-wt$.

(D) The fundamental frequency $f$ of a stretched string is given by the formula: $f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$.
Here,$T$ is the tension and $\mu$ is the linear mass density of the string.
Since $T$ and $\mu$ remain constant,the product of frequency and length is constant: $f_1 l_1 = f_2 l_2$.
Given $f_1 = 512 \; Hz$,$l_1 = 0.5 \; m$,and $f_2 = 256 \; Hz$.
Substituting the values: $512 \times 0.5 = 256 \times l_2$.
$256 = 256 \times l_2$.
Therefore,$l_2 = 1 \; m$.

(B) For a sonometer wire,the frequency of vibration $n$ is given by $n = \frac{p}{2l} \sqrt{\frac{T}{\mu}}$,where $p$ is the number of loops (antinodes),$l$ is the length,$T$ is the tension,and $\mu$ is the linear mass density.
Since $\mu = \pi r^2 \rho$,the frequency formula becomes $n = \frac{p}{2l} \sqrt{\frac{T}{\pi r^2 \rho}}$.
For the composite wire,the frequency $n$ is the same for both parts $W_1$ and $W_2$. Also,$T$,$r$,and $l$ are the same for both wires.
Thus,$n_1 = n_2 \implies \frac{p_1}{2l} \sqrt{\frac{T}{\pi r^2 \rho_1}} = \frac{p_2}{2l} \sqrt{\frac{T}{\pi r^2 \rho_2}}$.
Simplifying,we get $\frac{p_1}{\sqrt{\rho_1}} = \frac{p_2}{\sqrt{\rho_2}}$.
Given $\rho_2 = 4\rho_1$,we have $\frac{p_1}{\sqrt{\rho_1}} = \frac{p_2}{\sqrt{4\rho_1}} = \frac{p_2}{2\sqrt{\rho_1}}$.
Therefore,$\frac{p_1}{p_2} = \frac{1}{2}$.

(B) The total length of the wire is $L = 110 \ cm = 1.1 \ m$. The ratio of the lengths is $6:3:2$. Let the lengths be $l_1, l_2, l_3$.
Sum of parts $= 6+3+2 = 11$.
$l_1 = (6/11) \times 110 = 60 \ cm = 0.6 \ m$.
$l_2 = (3/11) \times 110 = 30 \ cm = 0.3 \ m$.
$l_3 = (2/11) \times 110 = 20 \ cm = 0.2 \ m$.
The fundamental frequency of a wire is given by $f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $T = 400 \ N$ and $\mu = 0.01 \ kg/m$.
$\sqrt{\frac{T}{\mu}} = \sqrt{\frac{400}{0.01}} = \sqrt{40000} = 200 \ m/s$.
$f_1 = \frac{200}{2 \times 0.6} = \frac{100}{0.6} = \frac{1000}{6} \ Hz$.
$f_2 = \frac{200}{2 \times 0.3} = \frac{100}{0.3} = \frac{1000}{3} \ Hz$.
$f_3 = \frac{200}{2 \times 0.2} = \frac{100}{0.2} = 500 \ Hz = \frac{1000}{2} \ Hz$.
The common frequency is the Least Common Multiple $(LCM)$ of the frequencies.
$LCM(\frac{1000}{6}, \frac{1000}{3}, \frac{1000}{2}) = \frac{LCM(1000, 1000, 1000)}{GCD(6, 3, 2)} = \frac{1000}{1} = 1000 \ Hz$.

(D) Total length of the wire,$L = 114\, cm$.
The fundamental frequency $n$ of a stretched string is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,which implies $n \propto \frac{1}{L}$.
Given the ratio of frequencies $n_1 : n_2 : n_3 = 1 : 3 : 4$.
Therefore,the ratio of the lengths of the three segments is $L_1 : L_2 : L_3 = \frac{1}{1} : \frac{1}{3} : \frac{1}{4} = 12 : 4 : 3$.
The sum of the ratio parts is $12 + 4 + 3 = 19$.
Calculating the lengths:
$L_1 = \frac{12}{19} \times 114 = 12 \times 6 = 72\, cm$.
$L_2 = \frac{4}{19} \times 114 = 4 \times 6 = 24\, cm$.
$L_3 = \frac{3}{19} \times 114 = 3 \times 6 = 18\, cm$.
To divide the wire into these segments,the first bridge should be placed at $72\, cm$ from one end,and the second bridge should be placed at $72 + 24 = 96\, cm$ from the same end.

(D) The linear mass density $\mu = \frac{m}{L} = \frac{5 \times 10^{-3} \, kg}{1 \, m} = 5 \times 10^{-3} \, kg/m$.
The wave speed $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{8.0}{5 \times 10^{-3}}} = \sqrt{1600} = 40 \, m/s$.
The wavelength $\lambda = \frac{v}{f} = \frac{40 \, m/s}{100 \, Hz} = 0.4 \, m$.
The separation between successive nodes is equal to $\frac{\lambda}{2}$.
Separation $= \frac{0.4 \, m}{2} = 0.2 \, m = 20 \, cm$.

(A) Let the mass per unit length of wires $A$ and $B$ be $\mu_1$ and $\mu_2$ respectively.
Since the materials are the same,the density $\rho$ is the same.
$\mu_1 = \rho \pi r^2$ and $\mu_2 = \rho \pi (2r)^2 = 4 \rho \pi r^2 = 4 \mu_1$.
The tension $T$ is the same in both wires.
The wave speed in the wires is $v = \sqrt{T/\mu}$.
Thus,$v_1 = \sqrt{T/\mu_1} = v$ and $v_2 = \sqrt{T/\mu_2} = \sqrt{T/(4\mu_1)} = v/2$.
For a wire of length $L$ fixed at both ends,the frequency of the $n$-th harmonic is $f = n \frac{v}{2L}$.
Since the joint is a node,both wires vibrate independently with the same frequency $f$.
$f = p \frac{v_1}{2L} = q \frac{v_2}{2L}$,where $p$ and $q$ are the number of loops (which equals the number of antinodes).
$p \frac{v}{2L} = q \frac{v/2}{2L} \implies p = q/2 \implies p/q = 1/2$.
Therefore,the ratio $p : q$ is $1 : 2$.

(B) For a string clamped at both ends,the length $\ell$ in the $n^{th}$ harmonic is given by $\ell = n \frac{\lambda}{2}$.
Given $n = 4$,we have $\ell = 4 \frac{\lambda}{2} = 2\lambda$.
The general equation of a stationary wave is $Y = A \sin(kx) \cos(\omega t)$,where $k = \frac{2\pi}{\lambda}$.
Comparing the given equation $Y = 0.3 \sin(0.157 x) \cos(200\pi t)$ with the general form,we get $k = 0.157$.
Since $0.157 \approx \frac{\pi}{20}$,we have $\frac{2\pi}{\lambda} = \frac{\pi}{20}$.
Solving for $\lambda$,we get $\lambda = 40 \, m$.
Substituting this into the length formula: $\ell = 2 \lambda = 2 \times 40 = 80 \, m$.

(D) The frequency of the $n^{th}$ harmonic for a string fixed at both ends is given by $f_n = \frac{n v}{2L}$.
Given: length $L = 2.0\, m$,frequency $f_3 = 240\, Hz$,and harmonic $n = 3$.
Substituting the values into the formula: $240 = \frac{3 \times v}{2 \times 2.0}$.
$240 = \frac{3v}{4} \Rightarrow 3v = 960 \Rightarrow v = 320\, m/s$.
The fundamental frequency $(n=1)$ is $f_1 = \frac{v}{2L} = \frac{320}{2 \times 2.0} = \frac{320}{4} = 80\, Hz$.

(B) The fundamental frequency $n$ of a stretched wire is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.
Since stress $\sigma = \frac{T}{A}$ and mass per unit length $\mu = \frac{m}{L} = \frac{\rho \cdot A \cdot L}{L} = \rho A$,we have $\frac{T}{\mu} = \frac{\sigma A}{\rho A} = \frac{\sigma}{\rho}$.
Substituting this into the frequency formula: $n = \frac{1}{2L} \sqrt{\frac{\sigma}{\rho}}$.
Given: $L = 0.75\, m$,$\sigma = 8.1 \times 10^8\, N/m^2$,and $\rho = 9 \times 10^3\, kg/m^3$.
$n = \frac{1}{2 \times 0.75} \sqrt{\frac{8.1 \times 10^8}{9 \times 10^3}} = \frac{1}{1.5} \sqrt{0.9 \times 10^5} = \frac{1}{1.5} \sqrt{9 \times 10^4} = \frac{300}{1.5} = 200\, Hz$.

(D) Let the frequency of the tuning fork be $n$.
The frequency of the sonometer wire $f$ is inversely proportional to its length $\ell$,i.e.,$f \propto \frac{1}{\ell}$.
Given that the tuning fork produces $5 \, Hz$ beats in both cases,the frequency of the wire is either $(n + 5)$ or $(n - 5)$.
Since frequency is inversely proportional to length,the higher frequency corresponds to the shorter length:
$n + 5 = \frac{k}{1}$
$n - 5 = \frac{k}{1.05}$
Dividing the two equations:
$\frac{n + 5}{n - 5} = \frac{1.05}{1}$
$n + 5 = 1.05(n - 5)$
$n + 5 = 1.05n - 5.25$
$0.05n = 10.25$
$n = \frac{10.25}{0.05} = 205 \, Hz$.

(A) The fundamental frequency of a stretched string is given by $v = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $T$ is tension and $\mu$ is mass per unit length.
Since $T$ and $\mu$ are constant for the string,$v \propto \frac{1}{l}$,which implies $vl = k$ (a constant).
Therefore,for the three segments,we have $v_1 l_1 = v_2 l_2 = v_3 l_3 = k$.
This gives $l_1 = \frac{k}{v_1}$,$l_2 = \frac{k}{v_2}$,and $l_3 = \frac{k}{v_3}$.
The original length $l$ of the string is the sum of the lengths of the segments: $l = l_1 + l_2 + l_3$.
Substituting the expressions for lengths in terms of frequencies: $\frac{k}{v} = \frac{k}{v_1} + \frac{k}{v_2} + \frac{k}{v_3}$.
Dividing both sides by $k$,we get $\frac{1}{v} = \frac{1}{v_1} + \frac{1}{v_2} + \frac{1}{v_3}$.

(D) The fundamental frequency of a stretched string is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $L$ is the length,$T$ is the tension,and $\mu$ is the mass per unit length.
Since $\mu$ remains constant,$n \propto \frac{1}{L} \sqrt{T}$.
Let the initial length be $L_1$ and initial tension be $T_1$. The final length $L_2 = L_1 - 0.40L_1 = 0.60L_1$ and final tension $T_2 = T_1 + 0.44T_1 = 1.44T_1$.
The ratio of the final frequency $n_2$ to the initial frequency $n_1$ is given by $\frac{n_2}{n_1} = \frac{L_1}{L_2} \sqrt{\frac{T_2}{T_1}}$.
Substituting the values: $\frac{n_2}{n_1} = \frac{L_1}{0.60L_1} \sqrt{\frac{1.44T_1}{T_1}} = \frac{1}{0.6} \times \sqrt{1.44} = \frac{1}{0.6} \times 1.2 = 2$.
Thus,the ratio is $2 : 1$.

(C) The fundamental frequency $f$ of a stretched wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $L$ is the length,$T$ is the tension,and $\mu$ is the linear mass density.
Since $\mu = \rho \times A = \rho \times \pi r^2$,where $\rho$ is the density of the material and $r$ is the radius,the formula becomes $f = \frac{1}{2L} \sqrt{\frac{T}{\rho \pi r^2}} = \frac{1}{2Lr} \sqrt{\frac{T}{\rho \pi}}$.
Given that $L$,$T$,and $\rho$ are the same for both wires,we have $f \propto \frac{1}{r}$.
Therefore,the ratio of the fundamental frequencies is $\frac{f_1}{f_2} = \frac{r_2}{r_1}$.
Given the ratio of radii $r_1 : r_2 = 1 : 2$,we have $\frac{r_2}{r_1} = \frac{2}{1}$.
Thus,the ratio of the fundamental frequencies is $2 : 1$.

(C) For a stretched string of constant tension $T$ and linear mass density $\mu$,the fundamental frequency $f$ is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.
Since $T$ and $\mu$ are constant for all segments,the frequency is inversely proportional to the length of the segment,i.e.,$f \propto \frac{1}{L}$.
Given lengths are $L_1 = 50\,cm$,$L_2 = 40\,cm$,and $L_3 = 10\,cm$.
The ratio of frequencies is $f_1 : f_2 : f_3 = \frac{1}{L_1} : \frac{1}{L_2} : \frac{1}{L_3}$.
Substituting the values: $f_1 : f_2 : f_3 = \frac{1}{50} : \frac{1}{40} : \frac{1}{10}$.
To simplify,multiply by the least common multiple of $50, 40, 10$,which is $200$.
$f_1 : f_2 : f_3 = \frac{200}{50} : \frac{200}{40} : \frac{200}{10} = 4 : 5 : 20$.

(A) The standard equation for a standing wave is $y(x,t) = A \sin(kx) \cos(\omega t)$.
Comparing this with the given equation $y(x,t) = 0.6 \sin \left( \frac{2\pi}{3}x \right) \cos (120\pi t)$,we get angular frequency $\omega = 120\pi \, rad/s$ and wave number $k = \frac{2\pi}{3} \, m^{-1}$.
The wave speed $v$ is given by $v = \frac{\omega}{k} = \frac{120\pi}{2\pi/3} = 180 \, m/s$.
The linear mass density $\mu$ is $\mu = \frac{M}{L} = \frac{3.0 \times 10^{-2} \, kg}{1.5 \, m} = 2 \times 10^{-2} \, kg/m$.
The wave speed in a string is related to tension $T$ by $v = \sqrt{\frac{T}{\mu}}$,so $T = \mu v^2$.
Substituting the values: $T = (2 \times 10^{-2} \, kg/m) \times (180 \, m/s)^2 = 2 \times 10^{-2} \times 32400 = 648 \, N$.

(D) The resonance frequencies of a string fixed at both ends are given by $f_n = n f_1$,where $f_1$ is the fundamental frequency.
The difference between two consecutive resonance frequencies is equal to the fundamental frequency:
$f_{n+1} - f_n = f_1 = 450 \, Hz - 375 \, Hz = 75 \, Hz$.
The fundamental frequency is $f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $L$ is the length of the string,$T$ is the tension,and $\mu$ is the linear mass density.
Substituting the values:
$75 = \frac{1}{2L} \sqrt{\frac{360}{4 \times 10^{-3}}}$
$75 = \frac{1}{2L} \sqrt{90000} = \frac{300}{2L}$
$L = \frac{300}{150} = 2 \, m$.
The total mass of the string $m$ is given by $m = \mu \times L$.
$m = (4 \times 10^{-3} \, kg/m) \times (2 \, m) = 8 \times 10^{-3} \, kg$.

(D) The fundamental frequency of a sonometer wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $\mu = \pi r^2 \rho$ is the linear mass density.
Substituting $\mu$,we get $n = \frac{1}{2l} \sqrt{\frac{T}{\pi r^2 \rho}} = \frac{1}{2lr} \sqrt{\frac{T}{\pi \rho}}$.
Since tension $T$ and density $\rho$ are constant,the frequency $n \propto \frac{1}{lr}$.
Given that the length is doubled $(l' = 2l)$ and the diameter is doubled,which implies the radius is also doubled $(r' = 2r)$.
The new frequency $n'$ is given by $n' = n \times (\frac{l}{l'}) \times (\frac{r}{r'}) = n \times (\frac{l}{2l}) \times (\frac{r}{2r}) = n \times \frac{1}{2} \times \frac{1}{2} = \frac{n}{4}$.

(B) For a string fixed at both ends vibrating in $n$ loops,the length of the string $L$ is given by the formula $L = \frac{n\lambda}{2}$.
Here,the number of loops $n = 3$.
The wavelength $\lambda = 10 \, cm$.
Substituting these values into the formula:
$L = \frac{3 \times 10}{2} = \frac{30}{2} = 15 \, cm$.
Therefore,the length of the string is $15 \, cm$.

(A) The resonant frequencies of a string fixed at both ends are given by $f_n = n \times f_1$,where $f_1 = \frac{v}{2L}$ is the fundamental frequency and $n = 1, 2, 3, \dots$ is an integer.
Given two consecutive resonant frequencies $f_n = 315\, Hz$ and $f_{n+1} = 420\, Hz$.
We know that the difference between two consecutive resonant frequencies is equal to the fundamental frequency $f_1$.
$f_1 = f_{n+1} - f_n = 420\, Hz - 315\, Hz = 105\, Hz$.
Alternatively,$\frac{f_{n+1}}{f_n} = \frac{n+1}{n} = \frac{420}{315} = \frac{4}{3}$.
This implies $n = 3$,so $f_3 = 315\, Hz$ and $f_4 = 420\, Hz$.
Since $f_3 = 3 \times f_1 = 315\, Hz$,we get $f_1 = \frac{315}{3} = 105\, Hz$.
The lowest resonant frequency is the fundamental frequency $f_1 = 105\, Hz$.

(D) The frequency of the $n$-th harmonic for a string fixed at both ends is given by $f_n = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$.
Given consecutive frequencies $f_n = 420 \; Hz$ and $f_{n+1} = 490 \; Hz$.
The difference between consecutive frequencies is the fundamental frequency $f_1 = f_{n+1} - f_n = 490 - 420 = 70 \; Hz$.
We know $f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}} = 70 \; Hz$.
Given $T = 540 \; N$ and $\mu = 6.0 \times 10^{-3} \; kg/m$.
Calculate the wave speed $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{540}{6.0 \times 10^{-3}}} = \sqrt{90,000} = 300 \; m/s$.
Substitute $v$ into the fundamental frequency equation: $70 = \frac{300}{2L}$.
Solving for $L$: $L = \frac{300}{140} \approx 2.14 \; m$.
Thus,the length $L$ is approximately $2.1 \; m$.

(N/A) Given:
Mass of the wire,$m = 3.5 \times 10^{-2} \;kg$
Linear mass density,$\mu = 4.0 \times 10^{-2} \;kg \;m^{-1}$
Frequency of vibration,$f = 45 \;Hz$
Length of the wire,$l = \frac{m}{\mu} = \frac{3.5 \times 10^{-2}}{4.0 \times 10^{-2}} = 0.875 \;m$
For the fundamental mode of vibration,the length of the wire $l$ is equal to half the wavelength,i.e.,$l = \frac{\lambda}{2}$.
Therefore,$\lambda = 2l = 2 \times 0.875 = 1.75 \;m$.
$(a)$ The speed of the transverse wave $(v)$ is given by:
$v = f \lambda = 45 \times 1.75 = 78.75 \;m/s$.
$(b)$ The tension $(T)$ in the string is given by the relation $v = \sqrt{\frac{T}{\mu}}$,so $T = v^2 \mu$.
$T = (78.75)^2 \times 4.0 \times 10^{-2} = 6201.5625 \times 0.04 = 248.06 \;N$.

(N/A) Consider a string of length $L$ fixed at both ends under tension $T$. Let $\mu$ be the linear mass density of the string.
The wave equation for a stationary wave is given by $y(x, t) = A \sin(kx) \cos(\omega t)$.
Since the string is fixed at $x = 0$ and $x = L$,the boundary conditions are $y(0, t) = 0$ and $y(L, t) = 0$.
Applying the boundary condition at $x = L$,we get $\sin(kL) = 0$,which implies $kL = n\pi$ for $n = 1, 2, 3, \dots$.
Since $k = \frac{2\pi}{\lambda}$,we have $\frac{2\pi L}{\lambda} = n\pi$,which gives $\lambda = \frac{2L}{n}$.
The speed of the wave in the string is $v = \sqrt{\frac{T}{\mu}}$.
Using the relation $v = f\lambda$,where $f$ is the frequency,we get $f = \frac{v}{\lambda}$.
Substituting $\lambda = \frac{2L}{n}$,we obtain $f = \frac{nv}{2L} = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$,where $n = 1, 2, 3, \dots$ represents the mode of vibration.

(B) The fundamental frequency of a sonometer wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $L$ is the length,$T$ is the tension,and $\mu$ is the linear mass density.
Initially,the wire vibrates in its fundamental mode (first harmonic) with frequency $f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}} = f$,where $f$ is the frequency of the tuning fork.
When the length is doubled $(L' = 2L)$ and tension $T$ remains constant,the new fundamental frequency becomes $f_1' = \frac{1}{2(2L)} \sqrt{\frac{T}{\mu}} = \frac{1}{2} f$.
For the tuning fork to still be in resonance,the wire must vibrate at a higher harmonic such that its frequency matches $f$. The $n$-th harmonic frequency is $f_n = n \times f_1'$.
Setting $f_n = f$,we get $n \times (\frac{1}{2} f) = f$,which implies $n = 2$.
Therefore,the tuning fork will be in resonance with the wire when the wire vibrates in its second harmonic (first overtone).

(N/A) For a string of length $L$ fixed at both ends,the frequency of the $n^{\text{th}}$ harmonic is given by the formula: $f_n = \frac{nv}{2L}$,where $n$ is the number of loops,$v$ is the speed of the transverse wave in the string,and $L$ is the length of the string.
Since $v$ and $L$ are constant for a given string under constant tension,the frequency $f_n$ is directly proportional to the number of loops $n$ $(f_n \propto n)$.
For $n=1, 2, 3, 4$,the frequencies are $f_1 = \frac{v}{2L}$,$f_2 = \frac{2v}{2L}$,$f_3 = \frac{3v}{2L}$,and $f_4 = \frac{4v}{2L}$.
Therefore,the ratio of the frequencies is $f_1 : f_2 : f_3 : f_4 = 1 : 2 : 3 : 4$.

(A) Given: Density $\rho = 9 \times 10^{-3} \,kg/cm^3 = 9 \times 10^{-3} / 10^{-6} \,kg/m^3 = 9000 \,kg/m^3$.
Length $L = 1 \,m$.
Strain $= 4.9 \times 10^{-4}$.
Young's modulus $Y = 9 \times 10^{10} \,N/m^2$.
For the lowest frequency (fundamental mode),$L = \lambda / 2$,so $\lambda = 2L = 2 \,m$.
The frequency $f$ is given by $f = v / \lambda = (1 / \lambda) \sqrt{T / \mu}$,where $T$ is tension and $\mu$ is mass per unit length.
Since $Y = (T/A) / \text{strain}$,we have $T = Y \cdot A \cdot \text{strain}$.
Also,$\mu = m/L = (\rho \cdot V) / L = (\rho \cdot A \cdot L) / L = \rho \cdot A$.
Substituting these into the frequency formula:
$f = (1 / 2L) \sqrt{(Y \cdot A \cdot \text{strain}) / (\rho \cdot A)} = (1 / 2L) \sqrt{(Y \cdot \text{strain}) / \rho}$.
Substituting the values:
$f = (1 / 2) \sqrt{(9 \times 10^{10} \times 4.9 \times 10^{-4}) / 9000} = (1 / 2) \sqrt{(9 \times 10^6 \times 4.9) / 9000} = (1 / 2) \sqrt{10^3 \times 4.9} = (1 / 2) \sqrt{4900} = 70 / 2 = 35 \,Hz$.

(C) The fundamental frequency $f$ of a stretched string is given by the formula: $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $L$ is the length,$T$ is the tension,and $\mu$ is the linear mass density.
Since the strings $X$ and $Z$ are identical and made of the same material,their length $L$ and linear mass density $\mu$ are the same.
Therefore,the frequency is directly proportional to the square root of the tension: $f \propto \sqrt{T}$.
Given $f_{x} = 450\, Hz$ and $f_{z} = 300\, Hz$,we have:
$\frac{f_{x}}{f_{z}} = \sqrt{\frac{T_{x}}{T_{z}}}$
Substituting the values:
$\frac{450}{300} = \sqrt{\frac{T_{x}}{T_{z}}}$
$1.5 = \sqrt{\frac{T_{x}}{T_{z}}}$
Squaring both sides:
$\frac{T_{x}}{T_{z}} = (1.5)^2 = 2.25$.