Transverse Stationary Waves and Sonometer Questions in English

(A) Let the lengths of the three segments be $L_1$,$L_2$,and $L_3$ respectively. The total length is $L_1 + L_2 + L_3 = 110 \ cm$.
For a sonometer wire under constant tension,the fundamental frequency $n$ is inversely proportional to the length $L$ $(n \propto 1/L)$,so $n_1 L_1 = n_2 L_2 = n_3 L_3 = k$.
Given the ratio of frequencies $n_1 : n_2 : n_3 = 1 : 2 : 3$,we have:
$n_1 = x, n_2 = 2x, n_3 = 3x$.
Substituting these into the relation:
$x L_1 = 2x L_2 = 3x L_3$.
This gives $L_2 = L_1/2$ and $L_3 = L_1/3$.
Substituting into the total length equation:
$L_1 + L_1/2 + L_1/3 = 110 \ cm$.
Multiplying by $6$ to clear denominators: $6 L_1 + 3 L_1 + 2 L_1 = 660 \ cm$.
$11 L_1 = 660 \ cm \Rightarrow L_1 = 60 \ cm$.
Then $L_2 = 30 \ cm$ and $L_3 = 20 \ cm$.
The first bridge is placed at $60 \ cm$ from end $A$. The second bridge is placed at $60 + 30 = 90 \ cm$ from end $A$.

(B) The frequency of the $p^{th}$ harmonic for a string fixed at both ends is given by $f_p = \frac{p}{2L} \sqrt{\frac{T}{m}}$.
Given $f_p = 320 \ Hz$ and the next higher frequency $f_{p+1} = 400 \ Hz$.
We have the ratio: $\frac{f_{p+1}}{f_p} = \frac{p+1}{p}$.
Substituting the values: $\frac{400}{320} = \frac{p+1}{p}$.
Simplifying the fraction: $\frac{5}{4} = \frac{p+1}{p}$.
Cross-multiplying gives: $5p = 4(p+1) \Rightarrow 5p = 4p + 4$.
Therefore,$p = 4$.

(C) For a string fixed at both ends,the frequency of the $n$-th harmonic is given by $f_n = n \cdot f_1$,where $f_1$ is the fundamental frequency.
The $n$-th harmonic corresponds to $(n-1)$-th overtone.
Therefore,the third overtone corresponds to the $4$-th harmonic $(n = 4)$.
In the $n$-th harmonic,the number of loops is $n$.
For the $4$-th harmonic,there are $4$ loops.
The number of nodes is $(n + 1) = 4 + 1 = 5$.
The number of antinodes is $n = 4$.
Thus,there are $5$ nodes and $4$ antinodes.

(D) The difference between two consecutive resonant frequencies of a stretched string is equal to the fundamental frequency $(f_0)$.
$f_0 = f_2 - f_1 = 450 \ Hz - 375 \ Hz = 75 \ Hz$.
The formula for the fundamental frequency is $f_0 = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $T = 180 \ N$ and $\mu = 2 \times 10^{-3} \ kg/m$.
Substituting the values: $75 = \frac{1}{2L} \sqrt{\frac{180}{2 \times 10^{-3}}} = \frac{1}{2L} \sqrt{90000} = \frac{300}{2L} = \frac{150}{L}$.
Thus,$L = \frac{150}{75} = 2 \ m$.
The total mass of the string is $m = \mu \times L = (2 \times 10^{-3} \ kg/m) \times (2 \ m) = 4 \times 10^{-3} \ kg = 4 \ g$.

(D) Given: $T_A = T_B = T$,$r_A = 2r_B$.
The frequency of the $n$-th overtone for a string is given by $f_n = (n+1) \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $\mu = \pi r^2 \rho$.
First overtone of $A$ $(n=1)$: $f_{A,1} = 2 \times \frac{1}{2L_A} \sqrt{\frac{T}{\pi r_A^2 \rho}} = \frac{1}{L_A r_A} \sqrt{\frac{T}{\pi \rho}}$.
Second overtone of $B$ $(n=2)$: $f_{B,2} = 3 \times \frac{1}{2L_B} \sqrt{\frac{T}{\pi r_B^2 \rho}} = \frac{3}{2L_B r_B} \sqrt{\frac{T}{\pi \rho}}$.
Equating the two frequencies: $\frac{1}{L_A r_A} = \frac{3}{2L_B r_B}$.
Substituting $r_A = 2r_B$: $\frac{1}{L_A (2r_B)} = \frac{3}{2L_B r_B}$.
Simplifying: $\frac{1}{2L_A} = \frac{3}{2L_B} \implies \frac{L_B}{L_A} = \frac{3}{1}$.

(A) Let the initial length and tension be $l$ and $T$ respectively. The mass per unit length $m$ remains constant.
The fundamental frequency is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$.
After the changes:
New length $l' = l - 0.40l = 0.6l = \frac{3}{5}l$.
New tension $T' = T + 0.44T = 1.44T = \frac{144}{100}T$.
The new frequency $n'$ is given by $n' = \frac{1}{2l'} \sqrt{\frac{T'}{m}}$.
Taking the ratio of final to initial frequency:
$\frac{n'}{n} = \frac{l}{l'} \sqrt{\frac{T'}{T}} = \frac{l}{0.6l} \sqrt{\frac{1.44T}{T}} = \frac{1}{0.6} \times \sqrt{1.44} = \frac{1}{0.6} \times 1.2 = \frac{1.2}{0.6} = 2$.
Therefore,the ratio of final to initial frequency is $2:1$.

(A) The fundamental frequency of a stretched wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $\mu = \pi r^2 \rho$.
Thus,$n = \frac{1}{2lr} \sqrt{\frac{T}{\pi \rho}}$.
The first overtone of the first wire is $n_1 = 2 \times n_{1, \text{fund}} = 2 \times \frac{1}{2l_1 r_1} \sqrt{\frac{T}{\pi \rho}} = \frac{1}{l_1 r_1} \sqrt{\frac{T}{\pi \rho}}$.
The second overtone of the second wire is $n_2 = 3 \times n_{2, \text{fund}} = 3 \times \frac{1}{2l_2 r_2} \sqrt{\frac{T}{\pi \rho}} = \frac{3}{2l_2 r_2} \sqrt{\frac{T}{\pi \rho}}$.
Given $n_1 = n_2$,we have $\frac{1}{l_1 r_1} = \frac{3}{2l_2 r_2}$.
Rearranging for the ratio of lengths: $\frac{l_1}{l_2} = \frac{2r_2}{3r_1}$.
Since $r_1 = 2r_2$,substituting this gives $\frac{l_1}{l_2} = \frac{2r_2}{3(2r_2)} = \frac{1}{3}$.

(C) The fundamental frequency of a sonometer wire is given by $f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $\mu = \text{mass per unit length} = \rho \cdot A = \rho \cdot \pi \cdot (\frac{d}{2})^2 = \frac{\rho \pi d^2}{4}$.
Substituting $\mu$,we get $f = \frac{1}{2l} \sqrt{\frac{T}{\frac{\rho \pi d^2}{4}}} = \frac{1}{ld} \sqrt{\frac{T}{\pi \rho}}$.
For wire $A$: $n = \frac{1}{ld} \sqrt{\frac{T}{\pi \rho_1}}$.
For wire $B$: $n = \frac{1}{l(2d)} \sqrt{\frac{2T}{\pi \rho_2}}$.
Equating the frequencies: $\frac{1}{ld} \sqrt{\frac{T}{\pi \rho_1}} = \frac{1}{2ld} \sqrt{\frac{2T}{\pi \rho_2}}$.
Canceling common terms: $\sqrt{\frac{1}{\rho_1}} = \frac{1}{2} \sqrt{\frac{2}{\rho_2}}$.
Squaring both sides: $\frac{1}{\rho_1} = \frac{1}{4} \cdot \frac{2}{\rho_2} = \frac{1}{2\rho_2}$.
Therefore,$2\rho_2 = \rho_1$,which gives $\rho_2 = \frac{\rho_1}{2}$.

(C) The fundamental frequency of a vibrating string 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.
Mass per unit length $m = \frac{\text{Volume} \times \text{density}}{l} = \frac{(\pi R^2 l) \rho}{l} = \pi R^2 \rho$.
Since $\pi$ and $\rho$ are constant,$m \propto R^2$.
Substituting this into the frequency formula,we get $n \propto \frac{1}{l \sqrt{R^2}} \propto \frac{1}{lR}$.
Given $l_2 = 2l_1$ and $R_2 = 2R_1$,the ratio of the new frequency $n_2$ to the original frequency $n_1$ is:
$\frac{n_2}{n_1} = \frac{l_1 R_1}{l_2 R_2} = \frac{l_1 R_1}{(2l_1)(2R_1)} = \frac{1}{4}$.
Therefore,the new frequency $n_2 = \frac{n}{4}$.

(A) The fundamental frequency of a sonometer wire is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension in the wire.
Since the wire carries a block of mass $M$,the tension $T = Mg = V\rho g$,where $V$ is the volume of the block.
Thus,$n \propto \sqrt{T} \propto \sqrt{\rho Vg}$.
When the block is immersed in a liquid of density $\sigma$,the effective weight (tension) becomes $T' = V(\rho - \sigma)g$.
The new frequency $n'$ is given by $n' \propto \sqrt{T'}$.
Taking the ratio: $\frac{n'}{n} = \sqrt{\frac{T'}{T}} = \sqrt{\frac{V(\rho - \sigma)g}{V\rho g}} = \sqrt{\frac{\rho - \sigma}{\rho}}$.
Therefore,$n' = n \left[\frac{\rho - \sigma}{\rho}\right]^{\frac{1}{2}}$.

(C) The resonant frequencies of a string fixed at both ends are given by $f_n = n f_0$,where $f_0$ is the fundamental frequency (lowest resonant frequency) and $n = 1, 2, 3, \dots$.
Given that there are no resonant frequencies between $315 \,Hz$ and $420 \,Hz$,these two frequencies must be consecutive harmonics,i.e.,$f_n = 315 \,Hz$ and $f_{n+1} = 420 \,Hz$.
Thus,$n f_0 = 315$ and $(n+1) f_0 = 420$.
Subtracting the two equations: $(n+1) f_0 - n f_0 = 420 - 315$,which gives $f_0 = 105 \,Hz$.
Therefore,the lowest resonant frequency is $105 \,Hz$.

(A) The fundamental frequency $n$ of a sonometer wire is given by the formula:
$n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$ ---$(1)$
where $T$ is the tension and $\mu$ is the linear mass density.
When the tension is increased by $8 \,N$, the new tension becomes $(T + 8) \,N$ and the new frequency becomes $3n$. Thus:
$3n = \frac{1}{2l} \sqrt{\frac{T + 8}{\mu}}$ ---$(2)$
Dividing equation $(2)$ by equation $(1)$, we get:
$\frac{3n}{n} = \frac{\frac{1}{2l} \sqrt{\frac{T + 8}{\mu}}}{\frac{1}{2l} \sqrt{\frac{T}{\mu}}}$
$3 = \sqrt{\frac{T + 8}{T}}$
Squaring both sides:
$9 = \frac{T + 8}{T}$
$9T = T + 8$
$8T = 8$
$T = 1 \,N$
Therefore, the initial tension applied to the wire was $1 \,N$.

(A) The fundamental frequency $f$ of a stretched string is given by $f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$.
Since the tension $T$ and mass per unit length $\mu$ are constant, we have $f \propto \frac{1}{l}$, which implies $f_1 l_1 = f_2 l_2$.
Given $f_1 = 800 \,Hz$, $l_1 = 50 \,cm$, and $f_2 = 1000 \,Hz$.
Substituting the values: $800 \times 50 = 1000 \times l_2$.
$l_2 = \frac{800 \times 50}{1000} = 40 \,cm$.
The change in length is $\Delta l = l_1 - l_2 = 50 \,cm - 40 \,cm = 10 \,cm$.

(C) The frequency of a sonometer wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the mass per unit length.
Since the wire is in unison with the same tuning fork,the frequency $f$ remains constant.
For the first case,tension $T_1 = w$,so $f = \frac{1}{2L_1} \sqrt{\frac{w}{\mu}}$.
For the second case,the weight is increased by $3w$,so the new tension $T_2 = w + 3w = 4w$. Thus,$f = \frac{1}{2L_2} \sqrt{\frac{4w}{\mu}}$.
Equating the two expressions for $f$:
$\frac{1}{2L_1} \sqrt{\frac{w}{\mu}} = \frac{1}{2L_2} \sqrt{\frac{4w}{\mu}}$
$\frac{1}{L_1} = \frac{1}{L_2} \sqrt{4}$
$\frac{1}{L_1} = \frac{2}{L_2}$
Therefore,$\frac{L_1}{L_2} = \frac{1}{2}$.

(B) The fundamental frequency $n$ of a stretched wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Since $l$ and $\mu$ are constant,$n \propto \sqrt{T}$.
An octave means the frequency is doubled,so $n_2 = 2n_1$.
Using the ratio: $\frac{n_2}{n_1} = \sqrt{\frac{T_2}{T_1}}$.
Substituting the values: $2 = \sqrt{\frac{T_2}{2 \ kgwt}}$.
Squaring both sides: $4 = \frac{T_2}{2 \ kgwt}$.
Therefore,$T_2 = 8 \ kgwt$.

(C) The frequency $f$ of a vibrating string is given by $f = \frac{p}{2L} \sqrt{\frac{T}{\mu}}$, where $p$ is the number of loops (antinodes), $L$ is the length, $T$ is the tension, and $\mu$ is the mass per unit length.
Since the frequency $f$, length $L$, and mass per unit length $\mu$ are constant, we have $p \propto \sqrt{T}$, which implies $T \propto p^2$.
Therefore, $T_1 p_1^2 = T_2 p_2^2$.
Given $T_1 = 1 \,kg$ (weight equivalent), $p_1 = 4$, and $p_2 = 2$.
Substituting the values: $1 \times (4)^2 = M \times (2)^2$.
$16 = 4M$.
$M = 4 \,kg$.

(B) The frequency of vibration $n$ must be the same for both segments of the wire.
The frequency of a stretched wire is given by $n = \frac{p}{2L} \sqrt{\frac{T}{\mu}}$,where $\mu = \pi r^2 \rho$ is the linear mass density.
For the first wire (radius $r$,length $L$): $n = \frac{p_1}{2L} \sqrt{\frac{T}{\pi r^2 \rho}}$.
For the second wire (radius $2r$,length $L$): $n = \frac{p_2}{2L} \sqrt{\frac{T}{\pi (2r)^2 \rho}} = \frac{p_2}{2L} \sqrt{\frac{T}{4 \pi r^2 \rho}} = \frac{p_2}{4L} \sqrt{\frac{T}{\pi r^2 \rho}}$.
Equating the frequencies: $\frac{p_1}{2L} \sqrt{\frac{T}{\pi r^2 \rho}} = \frac{p_2}{4L} \sqrt{\frac{T}{\pi r^2 \rho}}$.
Simplifying,we get $\frac{p_1}{2} = \frac{p_2}{4}$,which implies $\frac{p_1}{p_2} = \frac{2}{4} = \frac{1}{2}$.

(D) The frequency of a vibrating string is given by $f = \frac{p}{2L} \sqrt{\frac{T}{\mu}}$, where $p$ is the number of loops (antinodes), $L$ is the length, $T$ is the tension, and $\mu$ is the linear mass density.
Since the tuning fork frequency $f$ and length $L$ remain constant, we have $p \propto \sqrt{T}$, or $T \propto p^2$.
Therefore, $T_1 p_1^2 = T_2 p_2^2$.
Given $T_1 = 9 \,kg-wt$, $p_1 = 5$, and $p_2 = 3$.
Substituting the values: $9 \times 5^2 = M \times 3^2$.
$9 \times 25 = M \times 9$.
$M = 25 \,kg$.

(B) The frequency of vibration of a string is given by the formula:
$f = \frac{1}{2L} \sqrt{\frac{T}{m}}$
where $T$ is the tension,$L$ is the length,and $m$ is the mass per unit length.
Since $m = \pi r^2 \rho$ (where $r$ is the radius and $\rho$ is the density),the frequency $n$ is:
$n = \frac{1}{2L} \sqrt{\frac{T}{\pi r^2 \rho}} = \frac{1}{2Lr} \sqrt{\frac{T}{\pi \rho}}$
From this expression,we see that $n \propto \frac{1}{r}$ when $T$,$L$,and $\rho$ are constant.
If the radius is doubled $(r' = 2r)$,the new frequency $n'$ becomes:
$\frac{n'}{n} = \frac{r}{r'} = \frac{r}{2r} = \frac{1}{2}$
Therefore,$n' = \frac{n}{2}$.

(D) The frequency of a sonometer wire is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$. Since the tension $T$ and mass per unit length $\mu$ are constant, we have $n \propto \frac{1}{L}$, which implies $nL = \text{constant}$.
Let $n$ be the frequency of the tuning fork.
Initially, $n \times 25 = k$ (where $k$ is a constant).
When the length is decreased by $1 \,cm$, the new length is $24 \,cm$. The frequency increases to $(n + 6) \,Hz$.
Thus, $(n + 6) \times 24 = k$.
Equating the two expressions for $k$:
$25n = 24(n + 6)$
$25n = 24n + 144$
$n = 144 \,Hz$.

(A) The fundamental frequency $n$ of a stretched wire is given by the formula:
$n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$
where $L$ is the length,$T$ is the tension,and $\mu$ is the linear mass density.
Since $\mu = \text{Area} \times \text{density} = (\pi r^2) \rho$,we can write:
$n = \frac{1}{2L} \sqrt{\frac{T}{\pi r^2 \rho}} = \frac{1}{2Lr} \sqrt{\frac{T}{\pi \rho}}$
From this expression,we see that $n \propto \frac{1}{Lr}$.
Given that the length $L$ is doubled $(L_2 = 2L_1)$ and the diameter $D$ is doubled (which implies the radius $r$ is also doubled,$r_2 = 2r_1$):
$\frac{n_2}{n_1} = \frac{L_1 r_1}{L_2 r_2} = \frac{L_1 r_1}{(2L_1)(2r_1)} = \frac{1}{4}$
Therefore,$n_2 = \frac{n_1}{4} = \frac{n}{4}$.

(C) 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 $\mu = \text{Area} \times \text{Density} = (\pi R^2) \rho$.
Substituting $\mu$ into the frequency formula: $f = \frac{1}{2L} \sqrt{\frac{T}{\pi R^2 \rho}} = \frac{1}{2LR} \sqrt{\frac{T}{\pi \rho}}$.
Given that the material is the same,$\rho$ is constant. Since $T$ is also constant,we have $f \propto \frac{1}{LR}$.
For the first string: $L_1 = L$,$R_1 = 2r$. Thus,$f_1 \propto \frac{1}{L \cdot 2r} = \frac{1}{2Lr}$.
For the second string: $L_2 = 2L$,$R_2 = r$. Thus,$f_2 \propto \frac{1}{2L \cdot r} = \frac{1}{2Lr}$.
The ratio of their frequencies is $\frac{f_1}{f_2} = \frac{1/2Lr}{1/2Lr} = 1:1$.

(A) The frequency of a sonometer wire is given by the formula $n = \frac{p}{2L} \sqrt{\frac{T}{\mu}}$,where $p$ is the number of loops (antinodes),$L$ is the length,$T$ is the tension $(T = Mg)$,and $\mu$ is the linear mass density.
Since the frequency $n$,length $L$,and linear mass density $\mu$ remain constant,we have $p \propto \sqrt{T} \propto \sqrt{M}$.
For the first case: $p_1 = 5$ and $M_1 = 9 \ kg$.
For the second case: $p_2 = 3$ and $M_2 = m$.
Using the ratio: $\frac{p_1}{p_2} = \sqrt{\frac{M_1}{M_2}}$.
Substituting the values: $\frac{5}{3} = \sqrt{\frac{9}{m}}$.
Squaring both sides: $\frac{25}{9} = \frac{9}{m}$.
Solving for $m$: $m = \frac{9 \times 9}{25} = \frac{81}{25} = 3.24 \ kg$.
Wait,re-evaluating the relationship: $p_1 \sqrt{M_1} = p_2 \sqrt{M_2}$.
$5 \sqrt{9} = 3 \sqrt{m} \implies 5 \times 3 = 3 \sqrt{m} \implies 5 = \sqrt{m} \implies m = 25 \ kg$.

(A) The frequency of the $p$-th overtone for a string fixed at both ends is given by $f = (p+1) \frac{1}{2 \ell} \sqrt{\frac{T}{\mu}}$,where $\mu = \pi r^2 \rho$.
Since the material is the same,density $\rho$ is constant. The tension $T$ is also constant.
For the first wire,the first overtone $(p=1)$ is $f_1 = 2 \cdot \frac{1}{2 \ell_1} \sqrt{\frac{T}{\pi r_1^2 \rho}} = \frac{1}{\ell_1 r_1} \sqrt{\frac{T}{\pi \rho}}$.
For the second wire,the second overtone $(p=2)$ is $f_2 = 3 \cdot \frac{1}{2 \ell_2} \sqrt{\frac{T}{\pi r_2^2 \rho}} = \frac{3}{2 \ell_2 r_2} \sqrt{\frac{T}{\pi \rho}}$.
Given $f_1 = f_2$,we have $\frac{1}{\ell_1 r_1} = \frac{3}{2 \ell_2 r_2}$.
Given $r_1 = 2 r_2$,substituting this into the equation: $\frac{1}{\ell_1 (2 r_2)} = \frac{3}{2 \ell_2 r_2}$.
Simplifying,$\frac{1}{2 \ell_1} = \frac{3}{2 \ell_2}$,which gives $\frac{\ell_1}{\ell_2} = \frac{1}{3}$.

(C) For the $5^{\text{th}}$ harmonic, a string of length $L$ fixed at both ends vibrates in $5$ loops.
The relationship between length $L$ and wavelength $\lambda$ is given by $L = \frac{5\lambda}{2}$.
Comparing the given equation $Y = 3 \sin(0.4x) \cos(200\pi t)$ with the standard wave equation $Y = A \sin(kx) \cos(\omega t)$, we find the wave number $k = 0.4 \text{ rad/cm}$.
Since $k = \frac{2\pi}{\lambda}$, we have $\lambda = \frac{2\pi}{k} = \frac{2\pi}{0.4} = 5\pi \text{ cm}$.
Substituting the value of $\lambda$ into the length equation: $L = \frac{5 \times 5\pi}{2} = \frac{25\pi}{2} = 12.5\pi \text{ cm}$.

(C) The fundamental frequency of a stretched string is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $L$ is the length,$T$ is the tension,and $\mu$ is the mass per unit length.
Since the strings are identical,$\mu$ is the same for both.
For the first string: $N = \frac{1}{2\ell} \sqrt{\frac{T_1}{\mu}}$
For the second string: $1.5N = \frac{1}{2(2\ell)} \sqrt{\frac{T_2}{\mu}} = \frac{1}{4\ell} \sqrt{\frac{T_2}{\mu}}$
Dividing the two equations: $\frac{N}{1.5N} = \frac{\frac{1}{2\ell} \sqrt{\frac{T_1}{\mu}}}{\frac{1}{4\ell} \sqrt{\frac{T_2}{\mu}}}$
$\frac{1}{1.5} = \frac{4\ell}{2\ell} \sqrt{\frac{T_1}{T_2}}$
$\frac{1}{1.5} = 2 \sqrt{\frac{T_1}{T_2}}$
$\sqrt{\frac{T_1}{T_2}} = \frac{1}{1.5 \times 2} = \frac{1}{3}$
Squaring both sides: $\frac{T_1}{T_2} = \frac{1}{9}$.

(C) The fundamental frequency of a sonometer wire is given by the formula: $n = \frac{1}{2\ell} \sqrt{\frac{T}{m}}$,where $T$ is the tension,$\ell$ is the length,and $m$ is the mass per unit length.
Initially,$n = \frac{1}{2\ell} \sqrt{\frac{T}{m}}$.
When the tension is increased by $8 \ N$,the new tension is $T' = T + 8$. The new frequency is $3n$.
So,$3n = \frac{1}{2\ell} \sqrt{\frac{T+8}{m}}$.
Dividing the second equation by the first equation:
$\frac{3n}{n} = \frac{\frac{1}{2\ell} \sqrt{\frac{T+8}{m}}}{\frac{1}{2\ell} \sqrt{\frac{T}{m}}}$
$3 = \sqrt{\frac{T+8}{T}}$
Squaring both sides:
$9 = \frac{T+8}{T}$
$9T = T + 8$
$8T = 8$
$T = 1 \ N$.

(A) The frequency of a sonometer wire is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density. Since $L$ and $\mu$ are constant,$n \propto \sqrt{T}$.
In air,the tension $T_1 = mg = V \varrho g$,where $V$ is the volume of the load and $\varrho$ is its specific gravity (density relative to water).
When the load is immersed in water,the buoyant force acts on it. The effective tension becomes $T_2 = V(\varrho - 1)g$ (since the density of water is $1 \text{ g/cm}^3$).
Taking the ratio of the frequencies: $\frac{n_2}{n_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{V(\varrho - 1)g}{V \varrho g}} = \sqrt{\frac{\varrho - 1}{\varrho}}$.
Therefore,the new frequency $n_2 = n \sqrt{\frac{\varrho - 1}{\varrho}}$.

(B) The fundamental frequency of a sonometer wire is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.
Since tension $T$ and mass per unit length $\mu$ are constant,$n \propto \frac{1}{L}$.
Let the initial length be $L_1 = L$ and the new length be $L_2 = L + 0.25L = 1.25L$.
The initial frequency $n_1 = 50 \ Hz$.
The new fundamental frequency $n_2$ is given by $\frac{n_2}{n_1} = \frac{L_1}{L_2} = \frac{L}{1.25L} = \frac{1}{1.25} = 0.8$.
So,$n_2 = 0.8 \times 50 \ Hz = 40 \ Hz$.
The frequency of the second harmonic is $2n$. Initially,$2n_1 = 2 \times 50 = 100 \ Hz$. Finally,$2n_2 = 2 \times 40 = 80 \ Hz$.
The change in frequency is $100 \ Hz - 80 \ Hz = 20 \ Hz$.
The percentage decrease is $\frac{20}{100} \times 100 \% = 20 \%$.

(D) The fundamental frequency $n$ of a sonometer wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the mass per unit length.
Since $l$ and $\mu$ are constant,$n \propto \sqrt{T}$.
Let the initial tension be $T$ and the final tension be $T + 3$.
Given that the frequency becomes double,$n_2 = 2n_1$.
Using the ratio: $\frac{n_1}{n_2} = \sqrt{\frac{T_1}{T_2}}$.
Substituting the values: $\frac{1}{2} = \sqrt{\frac{T}{T+3}}$.
Squaring both sides: $\frac{1}{4} = \frac{T}{T+3}$.
Cross-multiplying: $T + 3 = 4T$.
$3 = 3T$,which gives $T = 1 \ kg$.

(D) The frequency of vibration $v$ of a sonometer wire is given by the formula:
$v = \frac{1}{2L} \sqrt{\frac{T}{m}}$
where $T$ is the tension (weight $w$) and $m$ is the mass per unit length.
For the first case,the frequency is:
$v = \frac{1}{2L_1} \sqrt{\frac{w}{m}} \quad (i)$
For the second case,the weight is reduced to $\frac{w}{4}$ and the length becomes $L_2$:
$v = \frac{1}{2L_2} \sqrt{\frac{w/4}{m}} = \frac{1}{2L_2} \cdot \frac{1}{2} \sqrt{\frac{w}{m}} = \frac{1}{4L_2} \sqrt{\frac{w}{m}} \quad (ii)$
Since the tuning fork frequency $v$ remains constant,we equate $(i)$ and $(ii)$:
$\frac{1}{2L_1} \sqrt{\frac{w}{m}} = \frac{1}{4L_2} \sqrt{\frac{w}{m}}$
$\frac{1}{2L_1} = \frac{1}{4L_2}$
$4L_2 = 2L_1$
$\frac{L_1}{L_2} = \frac{4}{2} = 2$
Therefore,the ratio $\frac{L_1}{L_2}$ is $2:1$.

(C) The fundamental frequency of a stretched string is given by $v = \frac{1}{2l} \sqrt{\frac{T}{m}}$.
Since $l$ and $m$ are constant,$v \propto \sqrt{T}$.
Let the initial frequency be $v$ and initial tension be $T$.
When tension is increased by $69 \%$,the new tension $T' = T + 0.69T = 1.69T$.
The new frequency is $v' = v + 9$.
Taking the ratio: $\frac{v'}{v} = \sqrt{\frac{T'}{T}} = \sqrt{\frac{1.69T}{T}} = \sqrt{1.69} = 1.3$.
Therefore,$v + 9 = 1.3v$.
$0.3v = 9$.
$v = \frac{9}{0.3} = 30 \ Hz$.

(A) The fundamental frequency of a stretched wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $\mu = \pi r^2 \rho$.
Thus,$f = \frac{1}{2Lr} \sqrt{\frac{T}{\pi \rho}}$.
The first overtone of the first wire is $f_1 = 2f_1 = \frac{2}{2L_1 r_1} \sqrt{\frac{T}{\pi \rho}} = \frac{1}{L_1 r_1} \sqrt{\frac{T}{\pi \rho}}$.
The second overtone of the second wire is $f_2 = 3f_2 = \frac{3}{2L_2 r_2} \sqrt{\frac{T}{\pi \rho}}$.
Given $f_1 = f_2$,we have $\frac{1}{L_1 r_1} = \frac{3}{2L_2 r_2}$.
Rearranging for the ratio of lengths: $\frac{L_1}{L_2} = \frac{2r_2}{3r_1}$.
Since $r_1 = 2r_2$,we substitute: $\frac{L_1}{L_2} = \frac{2r_2}{3(2r_2)} = \frac{2}{6} = \frac{1}{3}$.

(A) For a stretched string fixed at both ends,the frequency of the $ n $-th overtone is given by $ f_n = (n+1) f_1 $,where $ f_1 $ is the fundamental frequency.
For the second overtone,$ n = 2 $,so the string vibrates in the $ 3^{rd} $ harmonic mode.
In the $ 3^{rd} $ harmonic mode,the string has $ 3 $ loops.
The number of nodes $( N )$ in a standing wave with $ p $ loops is $ p+1 $. Here,$ p = 3 $,so the number of nodes is $ 3+1 = 4 $.
The number of antinodes $( A )$ is equal to the number of loops,which is $ 3 $.
Therefore,there are $ 4 $ nodes and $ 3 $ antinodes.

(A) The frequency of a vibrating string is given by the formula: $v = \frac{1}{2l} \sqrt{\frac{T}{m}}$,where $l$ is the length,$T$ is the tension,and $m$ is the linear mass density.
From this,we can see that $v \propto \frac{\sqrt{T}}{l}$.
Let the initial state be $v_1 = 200 \ Hz$,$l_1 = l$,and $T_1 = T$. Let the final state be $v_2 = 300 \ Hz$,$l_2 = 2l$,and $T_2 = T'$.
Taking the ratio: $\frac{v_2}{v_1} = \frac{\sqrt{T'}/l_2}{\sqrt{T}/l_1} = \sqrt{\frac{T'}{T}} \cdot \frac{l_1}{l_2}$.
Substituting the values: $\frac{300}{200} = \sqrt{\frac{T'}{T}} \cdot \frac{l}{2l}$.
$1.5 = \sqrt{\frac{T'}{T}} \cdot 0.5$.
$\sqrt{\frac{T'}{T}} = \frac{1.5}{0.5} = 3$.
Squaring both sides,we get $\frac{T'}{T} = 3^2 = 9$.
Therefore,the ratio of the new tension to the original tension is $9: 1$.

$(A)$ The fundamental frequency of a stretched wire is given by the formula: $f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$, where $\mu = \pi r^2 \rho$ is the linear mass density.
Substituting $\mu$, we get: $f = \frac{1}{2lr} \sqrt{\frac{T}{\pi \rho}}$.
Given the initial conditions: $f_1 = 600 \,Hz$, length $l_1 = l$, radius $r_1 = r$, and tension $T_1 = T$.
New conditions: $l_2 = 2l$, $r_2 = r/2$, and $T_2 = T/9$.
The ratio of frequencies is: $\frac{f_2}{f_1} = \frac{l_1}{l_2} \times \frac{r_1}{r_2} \times \sqrt{\frac{T_2}{T_1}}$.
Substituting the values: $\frac{f_2}{600} = \frac{l}{2l} \times \frac{r}{r/2} \times \sqrt{\frac{T/9}{T}}$.
$\frac{f_2}{600} = \frac{1}{2} \times 2 \times \sqrt{\frac{1}{9}} = 1 \times \frac{1}{3} = \frac{1}{3}$.
Therefore, $f_2 = \frac{600}{3} = 200 \,Hz$.

(A) The fundamental frequency of a stretched wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $\mu$ is the linear mass density.
Linear mass density $\mu = \text{mass per unit length} = \text{density} \times \text{cross-sectional area} = \rho \times (\pi \frac{D^2}{4})$.
Substituting $\mu$ into the frequency formula:
$f = \frac{1}{2L} \sqrt{\frac{T}{\rho \pi \frac{D^2}{4}}} = \frac{1}{2L} \sqrt{\frac{4T}{\pi \rho D^2}} = \frac{1}{2L} \cdot \frac{2}{D} \sqrt{\frac{T}{\pi \rho}} = \frac{1}{LD} \sqrt{\frac{T}{\pi \rho}}$.
Since $T$,$\rho$,and $\pi$ are constants,we have $f \propto \frac{1}{LD}$.

(C) Given: Length of wire $L = 1.25 \ m$,strain $\epsilon = 0.14 \% = 0.0014$,density $\rho = 7.7 \times 10^3 \ kg \ m^{-3}$,Young's modulus $Y = 2.2 \times 10^{11} \ N \ m^{-2}$.
Fundamental frequency $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $\mu = \rho A$.
Since $Y = \frac{T/A}{\epsilon}$,we have $\frac{T}{A} = Y \epsilon = 2.2 \times 10^{11} \times 0.0014 = 3.08 \times 10^8 \ N \ m^{-2}$.
Substituting $\frac{T}{\mu} = \frac{T}{\rho A} = \frac{Y \epsilon}{\rho}$ into the frequency formula:
$f = \frac{1}{2L} \sqrt{\frac{Y \epsilon}{\rho}} = \frac{1}{2 \times 1.25} \sqrt{\frac{3.08 \times 10^8}{7.7 \times 10^3}} = \frac{1}{2.5} \sqrt{0.4 \times 10^5} = \frac{1}{2.5} \sqrt{40000} = \frac{200}{2.5} = 80 \ Hz$.

(C) The fundamental frequency 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 $T$ and $\mu$ are constant,we have $f \propto \frac{1}{L}$,which implies $fL = \text{constant} = k$.
Therefore,$L = \frac{k}{f}$.
When the wire of length $L$ is divided into three segments of lengths $L_1, L_2, L_3$,we have $L = L_1 + L_2 + L_3$.
Substituting $L = \frac{k}{f}$,$L_1 = \frac{k}{f_1}$,$L_2 = \frac{k}{f_2}$,and $L_3 = \frac{k}{f_3}$ into the equation,we get:
$\frac{k}{f} = \frac{k}{f_1} + \frac{k}{f_2} + \frac{k}{f_3}$.
Dividing both sides by $k$,we obtain the relation: $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3}$.

(D) Mass per unit length of the given string is $\mu = \frac{m_{string}}{l} = \frac{2 \times 10^{-3} \ kg}{1 \ m} = 2 \times 10^{-3} \ kg \ m^{-1}$.
Speed of transverse waves on a string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension in the string. Since the body is suspended,$T = M_{body} \times g$.
Fundamental frequency for a string fixed at both ends is $f_0 = \frac{v}{2l} = \frac{1}{2l} \sqrt{\frac{M_{body} \times g}{\mu}}$.
Squaring both sides,we get $f_0^2 = \frac{M_{body} \times g}{4l^2 \mu}$.
Rearranging for the mass of the body: $M_{body} = \frac{4l^2 f_0^2 \mu}{g}$.
Substituting the values: $M_{body} = \frac{4 \times (1)^2 \times (100)^2 \times (2 \times 10^{-3})}{10} = \frac{4 \times 10000 \times 0.002}{10} = \frac{80}{10} = 8 \ kg$.

(C) The fundamental frequency $f$ of a stretched wire is given by the formula:
$f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$
where $T$ is the tension,$L$ is the length,and $\mu$ is the mass per unit length.
Given that the tension $T$ is the same for both wires,the ratio of the fundamental frequencies of wires $A$ and $B$ is:
$\frac{f_A}{f_B} = \frac{\frac{1}{2L_A} \sqrt{\frac{T}{\mu_A}}}{\frac{1}{2L_B} \sqrt{\frac{T}{\mu_B}}} = \frac{L_B}{L_A} \sqrt{\frac{\mu_B}{\mu_A}}$
We are given the ratio of lengths $\frac{L_A}{L_B} = \frac{1}{2}$,so $\frac{L_B}{L_A} = \frac{2}{1}$.
The mass per unit length $\mu$ is defined as $\mu = \frac{m}{L}$. Therefore,the ratio $\frac{\mu_A}{\mu_B}$ is:
$\frac{\mu_A}{\mu_B} = \frac{m_A / L_A}{m_B / L_B} = \left( \frac{m_A}{m_B} \right) \left( \frac{L_B}{L_A} \right)$
Given $\frac{m_A}{m_B} = \frac{2}{1}$ and $\frac{L_B}{L_A} = \frac{2}{1}$,we have:
$\frac{\mu_A}{\mu_B} = \left( \frac{2}{1} \right) \left( \frac{2}{1} \right) = \frac{4}{1}$
Thus,$\frac{\mu_B}{\mu_A} = \frac{1}{4}$.
Substituting these values into the frequency ratio formula:
$\frac{f_A}{f_B} = \left( \frac{2}{1} \right) \sqrt{\frac{1}{4}} = 2 \times \frac{1}{2} = 1$
Therefore,the ratio of the fundamental frequencies is $1: 1$.

(A) The fundamental frequency $f$ of a vibrating 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 string is the same,the tension $T$ and mass per unit length $\mu$ are constant for all segments.
Therefore,the fundamental frequency is inversely proportional to the length of the segment: $f \propto \frac{1}{l}$ or $l \propto \frac{1}{f}$.
Given the ratio of fundamental frequencies is $f_1 : f_2 : f_3 = 1 : 2 : 3$.
The ratio of the lengths of the segments is $l_1 : l_2 : l_3 = \frac{1}{f_1} : \frac{1}{f_2} : \frac{1}{f_3}$.
Substituting the given values,we get $l_1 : l_2 : l_3 = \frac{1}{1} : \frac{1}{2} : \frac{1}{3}$.
To simplify this ratio,multiply each term by the least common multiple of the denominators $(6)$: $l_1 : l_2 : l_3 = (1 \times 6) : (\frac{1}{2} \times 6) : (\frac{1}{3} \times 6) = 6 : 3 : 2$.

(D) The frequency of the fundamental tone of a stretched wire is given by $f = \frac{1}{2l} \sqrt{\frac{T}{m}}$,where $T$ is the tension,$l$ is the length,and $m$ is the mass per unit length.
Since $m = \text{Area} \times \text{density} = \frac{\pi d^2}{4} \rho$,we have $f = \frac{1}{2l} \sqrt{\frac{T}{\pi d^2 \rho / 4}} = \frac{1}{ld} \sqrt{\frac{T}{\pi \rho}}$.
Given that $T$,$l$,and $\rho$ are constant,we have $f \propto \frac{1}{d}$.
Therefore,$\frac{f_2}{f_1} = \frac{d_1}{d_2} = \frac{0.90}{0.93} \approx 0.9677$.
The percentage change in frequency is given by $\frac{f_2 - f_1}{f_1} \times 100 = \left( \frac{f_2}{f_1} - 1 \right) \times 100$.
Substituting the values: $\left( \frac{0.90}{0.93} - 1 \right) \times 100 = \left( \frac{0.90 - 0.93}{0.93} \right) \times 100 = \left( \frac{-0.03}{0.93} \right) \times 100 \approx -3.22 \%$.
Rounding to the nearest option,the percentage change is $-3.2 \%$.

(A) The frequency of a stretched string is given by $f = \frac{n}{2l} \sqrt{\frac{T}{m}}$,where $n$ is the harmonic number,$l$ is the length,$T$ is the tension,and $m$ is the mass per unit length.
Since $m = \pi r^2 \rho$ (where $\rho$ is the density),the frequency formula becomes $f = \frac{n}{2l r} \sqrt{\frac{T}{\pi \rho}}$.
For the first overtone of $A$,$n = 2$. Thus,$f_A = \frac{2}{2 l_A r_A} \sqrt{\frac{T}{\pi \rho}} = \frac{1}{l_A r_A} \sqrt{\frac{T}{\pi \rho}}$.
For the second overtone of $B$,$n = 3$. Thus,$f_B = \frac{3}{2 l_B r_B} \sqrt{\frac{T}{\pi \rho}}$.
Given $f_A = f_B$,we have $\frac{1}{l_A r_A} = \frac{3}{2 l_B r_B}$.
Rearranging for the ratio of lengths: $\frac{l_A}{l_B} = \frac{2 r_B}{3 r_A}$.
Given $r_A = 2 r_B$,we substitute to get $\frac{l_A}{l_B} = \frac{2 r_B}{3 (2 r_B)} = \frac{1}{3}$.

(C) $1$. Thermal strain produced due to change in temperature is $\Delta L / L = \alpha \Delta T$.
$2$. Thermal stress is $F / A = Y (\Delta L / L) = Y \alpha \Delta T$.
$3$. Tension $T = Y A \alpha \Delta T = (200 \times 10^9) \times (10^{-6}) \times (1.21 \times 10^{-5}) \times 20 = 48.4 \,N$.
$4$. Linear mass density $\mu = m / L = 0.1 / 1 = 0.1 \,kg/m$.
$5$. Wave speed $v = \sqrt{T / \mu} = \sqrt{48.4 / 0.1} = \sqrt{484} = 22 \,m/s$.
$6$. Fundamental frequency $f = v / (2L) = 22 / (2 \times 1) = 11 \,Hz$.

(C) The fundamental frequency of a stretched string is given by $f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $l$ is the length,$T$ is the tension,and $\mu$ is the mass per unit length.
Taking the ratio of initial frequency $f_1$ to final frequency $f_2$:
$\frac{f_1}{f_2} = \frac{l_2}{l_1} \sqrt{\frac{T_1}{T_2}}$
Given $l_2 = l_1(1 - \frac{x}{100})$,$T_2 = T_1(1 + \frac{44}{100}) = 1.44 T_1$,and $\frac{f_1}{f_2} = \frac{1}{2}$.
Substituting these values:
$\frac{1}{2} = \frac{l_1(1 - \frac{x}{100})}{l_1} \sqrt{\frac{T_1}{1.44 T_1}}$
$\frac{1}{2} = (1 - \frac{x}{100}) \times \frac{1}{\sqrt{1.44}}$
$\frac{1}{2} = (1 - \frac{x}{100}) \times \frac{1}{1.2}$
$0.6 = 1 - \frac{x}{100}$
$\frac{x}{100} = 0.4$
$x = 40$.

(A) The frequency of the $n$-th harmonic for a stretched string is given by $f_n = \frac{n}{2l} \sqrt{\frac{T}{\mu}}$,where $\mu = \pi r^2 \rho$.
For the first overtone of $A$ $(n=2)$: $f_{A} = \frac{2}{2l_A} \sqrt{\frac{T}{\pi r_A^2 \rho}} = \frac{1}{l_A r_A} \sqrt{\frac{T}{\pi \rho}}$.
For the second overtone of $B$ $(n=3)$: $f_{B} = \frac{3}{2l_B} \sqrt{\frac{T}{\pi r_B^2 \rho}} = \frac{3}{2l_B r_B} \sqrt{\frac{T}{\pi \rho}}$.
Given $f_A = f_B$ and $r_A = 2r_B$,we equate the expressions:
$\frac{1}{l_A r_A} = \frac{3}{2l_B r_B} \Rightarrow \frac{1}{l_A (2r_B)} = \frac{3}{2l_B r_B}$.
Simplifying,we get $\frac{1}{2l_A} = \frac{3}{2l_B} \Rightarrow \frac{l_A}{l_B} = \frac{1}{3}$.
Thus,the ratio of the lengths $l_A : l_B = 1:3$.

(D) The fundamental frequency of a vibrating wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $\mu$ is the mass per unit length.
Since the tension $T$ and the mass per unit length $\mu$ (as the wire is the same) remain constant,we have $n \propto \frac{1}{l}$.
Therefore,$\frac{n_1}{n_2} = \frac{l_2}{l_1}$.
Given $n_1 = 1 \ kHz$ and $n_2 = 1.001 \ kHz$,we have $\frac{l_2}{l_1} = \frac{1}{1.001}$.
Using the thermal expansion formula $l_2 = l_1(1 - \alpha \Delta t)$,where $\Delta t = 30^{\circ} C - 10^{\circ} C = 20^{\circ} C$:
$\frac{l_1}{1.001} = l_1(1 - \alpha \times 20)$.
$1 - 20\alpha = \frac{1}{1.001} \approx 1 - 0.001$.
$20\alpha = 0.001$.
$\alpha = \frac{0.001}{20} = 0.5 \times 10^{-4} /^{\circ} C$.