Transverse Stationary Waves and Sonometer Questions in English

(A) Let the frequency of the tuning fork be $n$ and the frequency of the sonometer wire be $f$. The frequency of a sonometer wire is inversely proportional to its length,i.e.,$f \propto \frac{1}{l}$.
Since $4$ beats per second are produced,the frequency of the wire is either $(n + 4)$ or $(n - 4)$.
As length increases,frequency decreases. Therefore,for $l_1 = 95 \ cm$,the frequency is $(n + 4)$,and for $l_2 = 100 \ cm$,the frequency is $(n - 4)$.
Using the relation $f_1 l_1 = f_2 l_2$,we have $(n + 4) \times 95 = (n - 4) \times 100$.
Dividing by $5$,we get $(n + 4) \times 19 = (n - 4) \times 20$.
$19n + 76 = 20n - 80$.
$n = 76 + 80 = 156 \ Hz$.

(A) Let the frequency of the tuning fork be $n$ and the frequency of the sonometer wire be $f$.
Given that the wire produces $4$ beats with the tuning fork,the frequency of the wire is $f = n \pm 4$.
Since frequency $f \propto \frac{1}{l}$,we have $f \cdot l = \text{constant}$.
For the initial length $l_1 = 50 \ cm$,$f_1 = n \pm 4$.
For the shortened length $l_2 = 49 \ cm$,$f_2 = n \pm 4$.
Since $l_2 < l_1$,the frequency $f_2 > f_1$.
Therefore,$f_1 = n - 4$ and $f_2 = n + 4$.
Using the relation $f_1 l_1 = f_2 l_2$:
$(n - 4) \times 50 = (n + 4) \times 49$
$50n - 200 = 49n + 196$
$n = 196 + 200 = 396 \ Hz$.

(C) When a string of length $l$ fixed at both ends vibrates in two segments,it forms two loops.
Each loop corresponds to half a wavelength,i.e.,$\frac{\lambda}{2}$.
Since there are two segments,the total length of the string is $l = 2 \times \frac{\lambda}{2}$.
Therefore,$l = \lambda$.
Thus,the wavelength of the corresponding wave is $l$.

(D) The fundamental frequency $n$ of a stretched string 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 the tension $T$ and mass per unit length $\mu$ remain constant,the frequency is inversely proportional to the length: $n \propto \frac{1}{l}$.
Therefore,$\frac{n_2}{n_1} = \frac{l_1}{l_2}$.
Given $l_1 = 1 \, cm$,$l_2 = \frac{1}{4} \, cm$,and $n_1 = 256 \, Hz$.
Substituting the values: $n_2 = n_1 \times \frac{l_1}{l_2} = 256 \times \frac{1}{1/4} = 256 \times 4 = 1024 \, Hz$.

(B) The length of the string $l = 1 \ m = 100 \ cm$. The mass $m = 5 \times 10^{-4} \ kg$. The tension $T = 20 \ N$.
When a string is plucked at a distance $x$ from one end,it vibrates in the $p^{th}$ harmonic such that $x = \frac{l}{2p}$.
Given $x = 25 \ cm$,we have $25 = \frac{100}{2p}$,which gives $2p = 4$,so $p = 2$.
The frequency of vibration $f$ is given by $f = \frac{p}{2l} \sqrt{\frac{T}{\mu}}$,where $\mu$ is the mass per unit length.
Here,$\mu = \frac{m}{l} = \frac{5 \times 10^{-4} \ kg}{1 \ m} = 5 \times 10^{-4} \ kg/m$.
Substituting the values: $f = \frac{2}{2 \times 1} \sqrt{\frac{20}{5 \times 10^{-4}}} = 1 \times \sqrt{4 \times 10^4} = 200 \ Hz$.

(B) The fundamental frequency of a sonometer wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$.
Since $l$ and $\mu$ are constant,$n \propto \sqrt{T}$,which implies $T \propto n^2$.
Taking the derivative,we get $\frac{\Delta T}{T} = 2 \frac{\Delta n}{n}$.
Given the initial frequency $n = 500 \, Hz$ and the beat frequency $\Delta n = 5 \, Hz$,the new frequency required is $n' = 505 \, Hz$.
The fractional change in frequency is $\frac{\Delta n}{n} = \frac{5}{500} = 0.01$ or $1 \%$.
Substituting this into the tension relation: $\frac{\Delta T}{T} \times 100 = 2 \times (\frac{\Delta n}{n} \times 100) = 2 \times 1 \% = 2 \%$.
Therefore,the tension must be increased by $2 \%$.

(A) The fundamental frequency $n$ of a sonometer wire is given by the formula $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$,where $l$ is the length,$T$ is the tension,and $m$ is the mass per unit length.
From this relation,we can see that $n \propto \sqrt{T}$.
Let the initial tension be $T_1$ and the initial frequency be $n_1$. Let the new tension be $T_2 = 4T_1$ and the new frequency be $n_2$.
Then,$\frac{n_2}{n_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{4T_1}{T_1}} = \sqrt{4} = 2$.
Therefore,the fundamental frequency increases by $2$ times.

(C) The frequency of vibration of a stretched string is given by the formula $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $n$ is the frequency,$L$ is the length,$T$ is the tension,and $\mu$ is the linear mass density.
From this relation,we can see that $n \propto \sqrt{T}$.
If the frequency is increased by a factor of two,the new frequency $n' = 2n$.
Therefore,$\frac{n'}{n} = \frac{\sqrt{T'}}{\sqrt{T}} = 2$.
Squaring both sides,we get $\frac{T'}{T} = 4$,which implies $T' = 4T$.
Thus,the tension must be made four times the original tension.

(D) 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 the wires have identical length,diameter,and material,$L$ and $\mu$ are constant for all wires.
Therefore,the frequency is directly proportional to the square root of the tension: $n \propto \sqrt{T}$.
Given the ratio of tensions $T_1 : T_2 : T_3 : T_4 = 1 : 4 : 9 : 16$,the ratio of their fundamental frequencies is:
$n_1 : n_2 : n_3 : n_4 = \sqrt{1} : \sqrt{4} : \sqrt{9} : \sqrt{16}$.
Calculating the square roots,we get $1 : 2 : 3 : 4$.
Thus,the correct ratio is $1 : 2 : 3 : 4$.

(C) Let the frequency of the tuning fork be $N$.
Since the frequency of a vibrating string is inversely proportional to its length $(f \propto 1/l)$,we have $f \cdot l = \text{constant}$.
For a sonometer wire,the frequency $f$ is given by $f = k/l$.
When the length is $20 \ cm$,the frequency is $f_1 = k/20$. Since it produces $5$ beats per second with the tuning fork,$f_1 = N \pm 5$.
When the length is $21 \ cm$,the frequency is $f_2 = k/21$. Since the beat frequency remains $5$,$f_2 = N \mp 5$.
As the length increases,the frequency decreases. Therefore,$f_1 = N + 5$ and $f_2 = N - 5$.
Thus,$(N + 5) \cdot 20 = (N - 5) \cdot 21$.
$20N + 100 = 21N - 105$.
$N = 205 \ Hz$.

(A) The frequency of the $n^{th}$ harmonic for a string fixed at both ends is given by $f_n = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$,where $n$ is the harmonic number.
For the $9^{th}$ harmonic,$f_9 = \frac{9}{2L} \sqrt{\frac{T}{\mu}}$.
For the $7^{th}$ harmonic,$f_7 = \frac{7}{2L} \sqrt{\frac{T}{\mu}}$.
Since $9 > 7$,it follows that $f_9 > f_7$.
Therefore,the frequency of the $9^{th}$ harmonic is higher than that of the $7^{th}$ harmonic.

(D) The fundamental frequency of a sonometer 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.
From this,we can see that $n \propto \frac{1}{l} \sqrt{T}$.
Let the initial frequency be $n$,initial length be $l$,and initial tension be $T$.
Given: New tension $T' = 4T$ and new length $l' = 2l$.
The new frequency $n'$ is given by:
$\frac{n'}{n} = \frac{l}{l'} \times \sqrt{\frac{T'}{T}}$
Substituting the values:
$\frac{n'}{n} = \frac{l}{2l} \times \sqrt{\frac{4T}{T}}$
$\frac{n'}{n} = \frac{1}{2} \times \sqrt{4} = \frac{1}{2} \times 2 = 1$
Therefore,$n' = n$.

(A) $Sonometer$ is a laboratory instrument used to investigate the relationship between the frequency of the sound produced by a plucked string and the tension,length,and mass per unit length of the string.
It works on the principle of resonance,where the frequency of the vibrating string is matched with the frequency of an external sound source.

(A) The fundamental frequency $n$ of a stretched string is inversely proportional to its length $l$,given by $n \propto \frac{1}{l}$.
This implies $n_1 l_1 = n_2 l_2$.
Given: $l_1 = 50 \ cm$,$n_1 = 270 \ Hz$,and $n_2 = 1000 \ Hz$.
We need to find $l_2$.
Using the relation $l_2 = l_1 \left( \frac{n_1}{n_2} \right)$,
$l_2 = 50 \times \left( \frac{270}{1000} \right)$,
$l_2 = 50 \times 0.27 = 13.5 \ cm$.

(C) The frequency $n$ of a stretched string is given by the formula $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Since $l$ and $\mu$ are constant,we have $n \propto \sqrt{T}$.
Therefore,$\frac{n_1}{n_2} = \sqrt{\frac{T_1}{T_2}}$.
Given $n_1 = n$,$n_2 = 2n$,and $T_1 = 10 \ N$.
Substituting these values: $\frac{n}{2n} = \sqrt{\frac{10}{T_2}}$.
$\frac{1}{2} = \sqrt{\frac{10}{T_2}}$.
Squaring both sides: $\frac{1}{4} = \frac{10}{T_2}$.
Thus,$T_2 = 40 \ N$.

(B) The fundamental frequency $n$ of a stretched string is given by the formula $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Since $L$ and $\mu$ are constant,we have $n \propto \sqrt{T}$.
Let the initial frequency be $n_1 = 100 \ Hz$ and the final frequency be $n_2 = 400 \ Hz$.
Then,$\frac{n_2}{n_1} = \sqrt{\frac{T_2}{T_1}}$.
Substituting the values: $\frac{400}{100} = \sqrt{\frac{T_2}{T_1}} \implies 4 = \sqrt{\frac{T_2}{T_1}}$.
Squaring both sides,we get $\frac{T_2}{T_1} = 4^2 = 16$.
Therefore,the tension must be increased by $16$ times.

(D) The fundamental frequency $n$ of a stretched 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.
From this formula,we have $n \propto \frac{\sqrt{T}}{l}$.
Let the initial frequency be $n_1$ and the final frequency be $n_2 = 2n_1$.
Let the initial length be $l_1$ and the final length be $l_2 = \frac{3}{4}l_1$.
Using the proportionality $n \propto \frac{\sqrt{T}}{l}$,we can write $\frac{n_2}{n_1} = \frac{\sqrt{T_2}/l_2}{\sqrt{T_1}/l_1} = \sqrt{\frac{T_2}{T_1}} \cdot \frac{l_1}{l_2}$.
Substituting the given values: $2 = \sqrt{\frac{T_2}{T_1}} \cdot \frac{l_1}{(3/4)l_1} = \sqrt{\frac{T_2}{T_1}} \cdot \frac{4}{3}$.
Squaring both sides: $4 = \frac{T_2}{T_1} \cdot \frac{16}{9}$.
Therefore,$\frac{T_2}{T_1} = 4 \cdot \frac{9}{16} = \frac{9}{4} = 2.25$.

(A) To generate the $p$-th harmonic in a string of length $l$,the string should be plucked at a distance $x = \frac{l}{2p}$ from one end to create an antinode,and touched at a node to suppress other harmonics.
For the second harmonic,$p = 2$.
Therefore,the plucking point is $x = \frac{l}{2 \times 2} = \frac{l}{4}$ from one end.
The string must be touched at a node to ensure the second harmonic mode is formed. In the second harmonic,there is a node at the center of the string,which is at distance $\frac{l}{2}$ from either end.
Thus,the string is plucked at $\frac{l}{4}$ and touched at $\frac{l}{2}$.

(A) The frequency of vibration of a sonometer wire 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 mass per unit length.
In the first case,$p_1 = 5$ and $T_1 = 9g$. Thus,$n = \frac{5}{2l} \sqrt{\frac{9g}{\mu}}$.
In the second case,$p_2 = 3$ and $T_2 = Mg$. Thus,$n = \frac{3}{2l} \sqrt{\frac{Mg}{\mu}}$.
Since the tuning fork is the same,the frequency $n$ remains constant. Equating the two expressions:
$\frac{5}{2l} \sqrt{\frac{9g}{\mu}} = \frac{3}{2l} \sqrt{\frac{Mg}{\mu}}$
Squaring both sides:
$25 \times 9g = 9 \times Mg$
$225g = 9Mg$
$M = \frac{225}{9} = 25 \ kg$.

(B) The frequency $n$ of a stretched string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension,$l$ is the length,and $\mu$ is the mass per unit length.
Since $n$ and $\mu$ are constant,we have $n \propto \frac{\sqrt{T}}{l}$,which implies $l \propto \sqrt{T}$.
Given that the tension is increased by $69\%$,the new tension $T_2 = T_1 + 0.69 T_1 = 1.69 T_1$.
Using the proportionality $l_2 / l_1 = \sqrt{T_2 / T_1}$,we get $l_2 / l_1 = \sqrt{1.69 T_1 / T_1} = \sqrt{1.69} = 1.3$.
Therefore,$l_2 = 1.3 l_1 = l_1 + 0.3 l_1$.
This represents an increase of $30\%$ in length.

(B) For a sonometer wire vibrating in its fundamental mode,the frequency $n$ is inversely proportional to the length $l$ of the wire,given by the relation $n \propto \frac{1}{l}$,or $nl = \text{constant}$.
Given:
Initial frequency $n_1 = 250 \ Hz$
Initial length $l_1 = 0.60 \ m$
Final length $l_2 = 0.40 \ m$
Using the relation $n_1 l_1 = n_2 l_2$:
$250 \times 0.60 = n_2 \times 0.40$
$n_2 = \frac{250 \times 0.60}{0.40}$
$n_2 = \frac{150}{0.40} = 375 \ Hz$.
Therefore,the frequency of the tuning fork is $375 \ Hz$.

(C) The fundamental frequency $f$ of a stretched string is given by $f = \frac{v}{2L}$,where $v$ is the wave speed and $L$ is the length of the string.
Since the tension and mass per unit length of the string remain constant,the wave speed $v$ is constant.
Therefore,$f \propto \frac{1}{L}$,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: $800 \times 50 = 1000 \times L_2$.
$L_2 = \frac{800 \times 50}{1000} = 40 \ cm$.
Thus,the required length of the string is $40 \ cm$.

(D) The frequency $n$ of a stretched string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}} = \frac{1}{2l} \sqrt{\frac{T}{\pi r^2 \rho}} = \frac{1}{l} \sqrt{\frac{T}{\pi d^2 \rho}}$,where $d$ is the diameter.
Taking the ratio of frequencies for two wires:
$\frac{n_1}{n_2} = \frac{l_2}{l_1} \sqrt{\frac{T_1}{T_2} \times \left(\frac{d_2}{d_1}\right)^2 \times \frac{\rho_2}{\rho_1}}$
Given ratios: $\frac{T_1}{T_2} = \frac{8}{1}$,$\frac{l_1}{l_2} = \frac{36}{35}$,$\frac{d_1}{d_2} = \frac{4}{1}$,$\frac{\rho_1}{\rho_2} = \frac{1}{2}$.
Substituting these values:
$\frac{n_1}{n_2} = \frac{35}{36} \sqrt{\frac{8}{1} \times \left(\frac{1}{4}\right)^2 \times \frac{2}{1}}$
$\frac{n_1}{n_2} = \frac{35}{36} \sqrt{8 \times \frac{1}{16} \times 2} = \frac{35}{36} \sqrt{1} = \frac{35}{36}$.
Since $n_1 = 360 \ Hz$ is the lower frequency,$n_1 = 360 \ Hz$ and $n_2 = 360 \times \frac{36}{35} = 370 \ Hz$.
Beat frequency = $|n_2 - n_1| = |370 - 360| = 10 \ Hz$.

(B) For a stretched wire fixed at both ends,the frequencies of harmonics are given by $n_k = k \cdot n_1$,where $n_1$ is the fundamental frequency (first harmonic) and $k = 1, 2, 3, \dots$.
The first overtone corresponds to the second harmonic $(k = 2)$.
Given that the first overtone $n_2 = 320 \ Hz$.
Since $n_2 = 2 \cdot n_1$,we have $320 \ Hz = 2 \cdot n_1$.
Therefore,the first harmonic $n_1 = \frac{320 \ Hz}{2} = 160 \ Hz$.

(D) The fundamental frequency of a stretched string is given by the formula: $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$,where $l$ is the length,$T$ is the tension,and $m$ is the mass per unit length.
From the formula,we can see that $n \propto \frac{1}{l} \sqrt{T}$.
Let the initial frequency be $n_1 = 200 \ Hz$,initial length be $l_1$,and initial tension be $T_1$.
Let the final frequency be $n_2$,final length be $l_2 = \frac{l_1}{4}$,and final tension be $T_2 = 4T_1$.
Using the proportionality: $\frac{n_2}{n_1} = \frac{l_1}{l_2} \sqrt{\frac{T_2}{T_1}}$.
Substituting the values: $\frac{n_2}{200} = \frac{l_1}{l_1/4} \sqrt{\frac{4T_1}{T_1}} = 4 \times \sqrt{4} = 4 \times 2 = 8$.
Therefore,$n_2 = 8 \times 200 = 1600 \ Hz$.

(B) The fundamental frequency of a stretched wire 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 the wires are similar,the mass per unit length $\mu$ and tension $T$ are the same for all segments.
Thus,$n \propto \frac{1}{l}$,which implies $nl = k$ (a constant).
For the three wires,we have $n_1 l_1 = n_2 l_2 = n_3 l_3 = k$.
This gives $l_1 = \frac{k}{n_1}$,$l_2 = \frac{k}{n_2}$,and $l_3 = \frac{k}{n_3}$.
The total length of the combined wire is $l = l_1 + l_2 + l_3$.
Substituting the values,we get $\frac{k}{n} = \frac{k}{n_1} + \frac{k}{n_2} + \frac{k}{n_3}$.
Dividing by $k$,we obtain $\frac{1}{n} = \frac{1}{n_1} + \frac{1}{n_2} + \frac{1}{n_3}$.

(A) When a rod is clamped at its mid-point,the mid-point acts as a node $(N)$ and the free ends act as antinodes $(A)$.
The length of the rod is $l = 100 \ cm = 1 \ m$.
For the fundamental mode of vibration,the length of the rod corresponds to half the wavelength,i.e.,$l = \frac{\lambda}{2}$,or $\lambda = 2l$.
The frequency $f$ is given by $f = \frac{v}{\lambda} = \frac{v}{2l}$.
Given $f = 2.53 \ kHz = 2530 \ Hz$ and $l = 1 \ m$,we have:
$2530 = \frac{v}{2 \times 1}$
$v = 2530 \times 2 = 5060 \ m/s$.
Converting to $km/s$,we get $v = 5.06 \ km/s$.

(D) The mass per unit length $\mu$ is given by $\mu = \frac{M}{L} = \frac{2 \times 10^{-4} \ kg}{0.5 \ m} = 4 \times 10^{-4} \ kg/m$.
The fundamental frequency $n_1$ of a stretched string is given by $n_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.
The frequency of the second harmonic $n_2$ is given by $n_2 = 2n_1 = 2 \times \frac{1}{2L} \sqrt{\frac{T}{\mu}} = \frac{1}{L} \sqrt{\frac{T}{\mu}}$.
Substituting the values: $n_2 = \frac{1}{0.5} \sqrt{\frac{20}{4 \times 10^{-4}}} = 2 \times \sqrt{5 \times 10^4} = 2 \times 223.6 = 447.2 \ Hz$.

(D) The fundamental frequency $n$ of a stretched string is given by the formula: $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$.
Since $l$ and $m$ are constant for a given string,the frequency is directly proportional to the square root of the tension: $n \propto \sqrt{T}$.
An octave corresponds to a frequency that is double the fundamental frequency,so $n' = 2n$.
Using the ratio: $\frac{n'}{n} = \sqrt{\frac{T'}{T}} = 2$.
Squaring both sides gives: $\frac{T'}{T} = 4$.
Therefore,the required tension is $T' = 4T = 4 \times 4 \ kg \ wt = 16 \ kg \ wt$.

(D) The fundamental frequency $n$ of a vibrating string is given by the formula: $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $\mu$ is the mass per unit length.
Since $\mu = \pi r^2 \rho$ (where $\rho$ is the density of the material),the frequency is $n = \frac{1}{2lr} \sqrt{\frac{T}{\pi \rho}}$.
Given that the material is the same,$\rho$ is constant. Since the tension $T$ is also the same,we have $n \propto \frac{1}{lr}$.
Therefore,the ratio of the frequencies is $\frac{n_1}{n_2} = \frac{l_2 r_2}{l_1 r_1}$.
Substituting the given values: $l_1 = L$,$l_2 = 2L$,$r_1 = 2r$,and $r_2 = r$.
$\frac{n_1}{n_2} = \frac{2L \times r}{L \times 2r} = \frac{2Lr}{2Lr} = 1$.

(C) 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 linear mass density.
Thus,$n \propto \sqrt{\frac{T}{r^2 \rho}}$.
Let the initial values be $T_1 = T$,$r_1 = r$,and $\rho_1 = \rho$. The new values are $T_2 = 2T$,$r_2 = 2r$,and $\rho_2 = \frac{\rho}{2}$.
The ratio of the frequencies is given by:
$\frac{n_2}{n_1} = \sqrt{\frac{T_2}{T_1} \cdot \left(\frac{r_1}{r_2}\right)^2 \cdot \frac{\rho_1}{\rho_2}}$
Substituting the values:
$\frac{n_2}{n_1} = \sqrt{\frac{2T}{T} \cdot \left(\frac{r}{2r}\right)^2 \cdot \frac{\rho}{\rho/2}} = \sqrt{2 \cdot \frac{1}{4} \cdot 2} = \sqrt{1} = 1$.
Therefore,$n_2 = n_1 = n$.

(A) The fundamental 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,$n \propto \sqrt{T}$.
Initially,the tension $T_1 = M g = 50.7 \times 10 = 507 \, N$.
When the mass is submerged in water,the buoyant force $F_B = V \rho_{water} g = 0.0075 \times 1000 \times 10 = 75 \, N$.
The new tension $T_2 = M g - F_B = 507 - 75 = 432 \, N$.
Using the ratio $\frac{n_1}{n_2} = \sqrt{\frac{T_1}{T_2}}$,we get $\frac{260}{n_2} = \sqrt{\frac{507}{432}}$.
$\frac{260}{n_2} = \sqrt{1.1736} \approx 1.0833$.
$n_2 = \frac{260}{1.0833} \approx 240 \, Hz$.

(C) For a string fixed at both ends,the wavelength $\lambda$ for the $p$-th harmonic is given by $\lambda = \frac{2L}{p}$,where $L$ is the length of the string and $p$ is the mode number.
Given: $L = 2 \ m$,$p = 4$,and frequency $f = 500 \ Hz$.
First,calculate the wavelength: $\lambda = \frac{2 \times 2}{4} = 1 \ m$.
The velocity of the wave $v$ is given by $v = f \times \lambda$.
Substituting the values: $v = 500 \ Hz \times 1 \ m = 500 \ m/s$.

(D) The fundamental frequency $n$ of a sonometer wire is given by the formula:
$n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$
where $T$ is the tension,$l$ is the length,and $\mu$ is the linear mass density.
Since $\mu = \pi r^2 \rho$ (where $r$ is the radius and $\rho$ is the density of the material),the formula becomes:
$n = \frac{1}{2l} \sqrt{\frac{T}{\pi r^2 \rho}} = \frac{1}{2l} \frac{1}{r} \sqrt{\frac{T}{\pi \rho}}$
This implies $n \propto \frac{\sqrt{T}}{r}$.
Given that the new radius $r_2 = 2r_1$ and the new tension $T_2 = \frac{T_1}{2}$,the ratio of the new frequency $n_2$ to the original frequency $n_1$ is:
$\frac{n_2}{n_1} = \frac{r_1}{r_2} \sqrt{\frac{T_2}{T_1}} = \frac{r_1}{2r_1} \sqrt{\frac{T_1/2}{T_1}} = \frac{1}{2} \times \sqrt{\frac{1}{2}} = \frac{1}{2\sqrt{2}}$.
Therefore,the new frequency $n_2 = \frac{n}{2\sqrt{2}}$.

(B) For a sonometer wire,the fundamental frequency $n$ is given by the relation:
$n = \frac{1}{2l} \sqrt{\frac{T}{m}}$
Since the tension $T$ and mass per unit length $m$ are constant,the frequency is inversely proportional to the resonating length $l$:
$n \propto \frac{1}{l}$
Therefore,$n_1 l_1 = n_2 l_2$
Given:
$n_1 = 256 \ Hz$,$l_1 = 25 \ cm$,$l_2 = 16 \ cm$
Substituting the values:
$256 \times 25 = n_2 \times 16$
$n_2 = \frac{256 \times 25}{16}$
$n_2 = 16 \times 25 = 400 \ Hz$
Thus,the frequency of the second tuning fork is $400 \ Hz$.

(B) In the condition of resonance,the frequency of the alternating current $(n)$ is equal to the fundamental frequency of the vibrating wire.
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 linear mass density.
Given:
Length $l = 1 \, m$
Tension $T = 10 \, kg \text{-} wt = 10 \times 9.8 \, N = 98 \, N$
Linear mass density $\mu = 9.8 \, g/m = 9.8 \times 10^{-3} \, kg/m$
Substituting the values:
$n = \frac{1}{2 \times 1} \sqrt{\frac{98}{9.8 \times 10^{-3}}}$
$n = \frac{1}{2} \sqrt{10000}$
$n = \frac{1}{2} \times 100 = 50 \, Hz$.

(B) The fundamental (lowest) frequency of a stretched wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$,where $T$ is the tension and $m$ is the mass per unit length.
Given: Density $\rho = 9 \times 10^3 \text{ kg/m}^3$,length $l = 1 \text{ m}$,extension $\Delta l = 4.9 \times 10^{-4} \text{ m}$,and Young's modulus $Y = 9 \times 10^{10} \text{ N/m}^2$.
Mass per unit length $m = \frac{M}{l} = \frac{A \cdot l \cdot \rho}{l} = A\rho$.
From Young's modulus,$Y = \frac{T/A}{\Delta l/l} \implies T = \frac{Y \cdot A \cdot \Delta l}{l}$.
Substituting $T$ and $m$ into the frequency formula:
$n = \frac{1}{2l} \sqrt{\frac{Y \cdot A \cdot \Delta l / l}{A \cdot \rho}} = \frac{1}{2l} \sqrt{\frac{Y \cdot \Delta l}{l \cdot \rho}}$.
Substituting the values:
$n = \frac{1}{2 \times 1} \sqrt{\frac{9 \times 10^{10} \times 4.9 \times 10^{-4}}{1 \times 9 \times 10^3}} = \frac{1}{2} \sqrt{\frac{9 \times 4.9 \times 10^6}{9 \times 10^3}} = \frac{1}{2} \sqrt{4.9 \times 10^3} = \frac{1}{2} \sqrt{4900} = \frac{70}{2} = 35 \text{ Hz}$.

(C) The frequency of vibration of a stretched string is given by $n = \frac{1}{2L} \sqrt{\frac{T}{m}}$.
Since the frequency $N$ remains constant,we have $L \propto \sqrt{T}$,where $T$ is the tension in the wire.
In air,the tension $T_{air} = V \rho g$,where $V$ is the volume of the stone and $\rho$ is its density.
When immersed in water,the tension becomes $T_{water} = V(\rho - \sigma)g$,where $\sigma$ is the density of water. Taking $\sigma = 1$ (specific gravity unit),$T_{water} = V(\rho - 1)g$.
Thus,$\frac{L}{l} = \sqrt{\frac{T_{air}}{T_{water}}} = \sqrt{\frac{V \rho g}{V(\rho - 1)g}} = \sqrt{\frac{\rho}{\rho - 1}}$.
Squaring both sides,$\frac{L^2}{l^2} = \frac{\rho}{\rho - 1}$.
$\rho(l^2) = L^2(\rho - 1) \implies \rho l^2 = L^2 \rho - L^2$.
$L^2 = \rho(L^2 - l^2)$.
Therefore,$\rho = \frac{L^2}{L^2 - l^2}$.

(C) The frequency of vibration of a string is given by $n = \frac{p}{2l} \sqrt{\frac{T}{m}}$,where $p$ is the number of loops,$l$ is the length,$T$ is the tension,and $m$ is the mass per unit length.
Since $n$,$l$,and $m$ are constant,we have $p \propto \frac{1}{\sqrt{T}}$,which implies $p_1 \sqrt{T_1} = p_2 \sqrt{T_2}$.
Given $p_1 = 4$ and $T_1 = (50 + 15) \,g = 65 \,g$.
For $p_2 = 6$,we have $4 \sqrt{65} = 6 \sqrt{T_2}$.
Squaring both sides: $16 \times 65 = 36 \times T_2$.
$T_2 = \frac{16 \times 65}{36} = \frac{4 \times 65}{9} \approx 28.89 \,g$.
The weight to be removed is $\Delta T = T_1 - T_2 = 65 - 28.89 = 36.11 \,g \approx 36 \,g$.
Converting to $kg$,we get $36 \,g = 0.036 \,kg \,wt$.

(A) The frequency $n$ of a vibrating wire is given by the formula:
$n = \frac{p}{2l} \sqrt{\frac{T}{\pi r^2 \rho}}$
where $p$ is the mode of vibration,$l$ is the length,$T$ is the tension,$r$ is the radius,and $\rho$ is the density.
Given that $l$,$r$,and $T$ are constant,the relationship between frequency $n$ and density $\rho$ is:
$n \propto \frac{1}{\sqrt{\rho}}$
This represents an inverse relationship,which graphically corresponds to a rectangular hyperbola. As $\sqrt{\rho}$ increases,$n$ decreases. Thus,the correct graph is a hyperbola.

(A) The frequency $n$ of a sonometer wire is inversely proportional to its length $l$,i.e.,$n \propto \frac{1}{l}$.
Given that the tuning fork produces $4$ beats per second with lengths $l_1 = 95 \, cm$ and $l_2 = 100 \, cm$.
Let the frequency of the tuning fork be $n$.
For $l_1 = 95 \, cm$,the frequency of the wire is $n_1 = \frac{k}{95}$. Since $l_1 < l_2$,$n_1 > n_2$. Thus,$n_1 = n + 4$.
For $l_2 = 100 \, cm$,the frequency of the wire is $n_2 = \frac{k}{100}$. Thus,$n_2 = n - 4$.
We have the relation: $(n + 4) \times 95 = (n - 4) \times 100$.
$95n + 380 = 100n - 400$.
$5n = 780$.
$n = 156 \, Hz$.

(C) The fundamental frequency of a stretched string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$.
Since the tension $T$ and mass per unit length $\mu$ are constant for the string,we have $n \propto \frac{1}{l}$,which implies $l = \frac{k}{n}$,where $k$ is a constant.
When the string is divided into three segments of lengths $l_{1}, l_{2}, l_{3}$ with fundamental frequencies $n_{1}, n_{2}, n_{3}$ respectively,we have $l_{1} = \frac{k}{n_{1}}$,$l_{2} = \frac{k}{n_{2}}$,and $l_{3} = \frac{k}{n_{3}}$.
The total length of the string is $l = l_{1} + l_{2} + l_{3}$.
Substituting the expressions for length in terms of frequency,we get $\frac{k}{n} = \frac{k}{n_{1}} + \frac{k}{n_{2}} + \frac{k}{n_{3}}$.
Dividing by $k$,we obtain the relation $\frac{1}{n} = \frac{1}{n_{1}} + \frac{1}{n_{2}} + \frac{1}{n_{3}}$.

(C) The fundamental frequency (first harmonic) is given as $n_1 = 100 \, Hz$.
In a string stretched between two fixed points,the harmonics are given by $n_k = k \times n_1$,where $k$ is the harmonic number $(k = 1, 2, 3, \dots)$.
The first overtone is the second harmonic $(n_2 = 2n_1)$.
The second overtone is the third harmonic $(n_3 = 3n_1)$.
The third overtone is the fourth harmonic $(n_4 = 4n_1)$.
Therefore,the frequency of the third overtone is $n_4 = 4 \times 100 \, Hz = 400 \, Hz$.

(D) When a string is plucked at a distance of $\frac{1}{n}$th of its length from one end,the $n$th harmonic is the most prominent one.
In this case,the string is plucked at $\frac{1}{4}$th of its length from the end,which means $n = 4$.
However,the question asks for the most prominent harmonic. If a string is plucked at $\frac{L}{n}$,the $n$th harmonic is absent if $n$ is a multiple of the harmonic number. Specifically,if plucked at $\frac{1}{4}$th,the $4$th harmonic has a node at the point of plucking and is therefore absent.
The harmonics present are those where the plucking point is an antinode. For a string plucked at $\frac{L}{4}$,the $1$st,$2$nd,$3$rd,$5$th,etc.,harmonics are present. The $2$nd harmonic is the most prominent among the lower harmonics because the plucking point coincides with an antinode of the $2$nd harmonic (which has antinodes at $\frac{L}{4}$ and $\frac{3L}{4}$). Thus,the $2$nd harmonic is the most prominent.

(C) For a vibrating string,the product of frequency $n$ and length $l$ is constant,i.e.,$n_1 l_1 = n_2 l_2 = n_3 l_3 = \dots = nl = k$ (constant).
Since the total length of the string is the sum of the lengths of its segments,we have $l_1 + l_2 + l_3 + \dots = l$.
Substituting $l_i = \frac{k}{n_i}$ and $l = \frac{k}{n}$ into the equation,we get $\frac{k}{n_1} + \frac{k}{n_2} + \frac{k}{n_3} + \dots = \frac{k}{n}$.
Dividing both sides by $k$,we obtain $\frac{1}{n} = \frac{1}{n_1} + \frac{1}{n_2} + \frac{1}{n_3} + \dots$.

(B) When the wire is cooled,it tends to contract,but since it is fixed between rigid supports,a tension $F$ is developed in the wire.
The thermal strain $\alpha \Delta T$ is equal to the elastic strain $\frac{F}{AY}$.
$\frac{F}{AY} = \alpha \Delta T \implies F = Y A \alpha \Delta T$
The frequency of the fundamental mode of transverse vibration is given by $f = \frac{1}{2L} \sqrt{\frac{F}{\mu}}$,where $\mu$ is the mass per unit length.
Since $\mu = \rho A$,we substitute $F$ and $\mu$ into the frequency formula:
$f = \frac{1}{2L} \sqrt{\frac{Y A \alpha \Delta T}{\rho A}} = \frac{1}{2L} \sqrt{\frac{Y \alpha \Delta T}{\rho}}$
Since $L, \Delta T$ are constants,$f \propto \sqrt{\frac{Y \alpha}{\rho}}$.

(D) 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 $f \propto \sqrt{T}$,we have $f \propto \sqrt{Mg}$.
When the mass $M$ is immersed in water,the effective tension is $T_w = Mg - V\rho_w g$,where $V$ is the volume of the mass and $\rho_w$ is the density of water. Given $f_w = f/2$,we have:
$\frac{f}{2} = \frac{1}{2l} \sqrt{\frac{Mg - V\rho_w g}{\mu}} \implies \frac{1}{4} = \frac{Mg - V\rho_w g}{Mg} = 1 - \frac{V\rho_w}{M}$.
Thus,$\frac{V\rho_w}{M} = 1 - \frac{1}{4} = \frac{3}{4}$,so $V = \frac{3M}{4\rho_w}$.
When the mass is immersed in a liquid of density $\rho_L$,the frequency is $f_L = f/3$. The effective tension is $T_L = Mg - V\rho_L g$. Thus:
$\frac{f}{3} = \frac{1}{2l} \sqrt{\frac{Mg - V\rho_L g}{\mu}} \implies \frac{1}{9} = \frac{Mg - V\rho_L g}{Mg} = 1 - \frac{V\rho_L}{M}$.
Substituting $V = \frac{3M}{4\rho_w}$:
$\frac{1}{9} = 1 - \frac{3M}{4\rho_w} \cdot \frac{\rho_L}{M} \implies \frac{3}{4} \cdot \frac{\rho_L}{\rho_w} = 1 - \frac{1}{9} = \frac{8}{9}$.
Therefore,the specific gravity $\frac{\rho_L}{\rho_w} = \frac{8}{9} \cdot \frac{4}{3} = \frac{32}{27}$.

(C) The fundamental frequency of a sonometer wire is given by $f = \frac{v}{2L} = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.
Since $\mu = A \rho$,we have $f = \frac{1}{2L} \sqrt{\frac{T}{A \rho}}$.
From Young's modulus $Y = \frac{T/A}{\Delta L/L}$,we get $\frac{T}{A} = Y \times \text{strain} = Y \times \frac{\Delta L}{L}$.
Substituting this into the frequency formula: $f = \frac{1}{2L} \sqrt{\frac{Y \times \text{strain}}{\rho}}$.
Given: $L = 1.5 \ m$,$\text{strain} = 1\% = 0.01$,$Y = 2.2 \times 10^{11} \ N/m^2$,$\rho = 7.7 \times 10^3 \ kg/m^3$.
$f = \frac{1}{2 \times 1.5} \sqrt{\frac{2.2 \times 10^{11} \times 0.01}{7.7 \times 10^3}}$.
$f = \frac{1}{3} \sqrt{\frac{2.2 \times 10^9}{7.7}} = \frac{1}{3} \sqrt{0.2857 \times 10^9} = \frac{1}{3} \sqrt{28.57 \times 10^7} \approx \frac{5345}{3} \approx 178.2 \ Hz$.

(C) The equation for the fundamental mode is $L = \frac{1}{2f\sqrt{\mu}} \sqrt{T}$.
Comparing this with the equation of a line $y = mx + c$,where $y = L$ and $x = \sqrt{T}$,the slope $m = \frac{1}{2f\sqrt{\mu}}$.
Given $m = 8.0 \times 10^{-3}$ and $\mu = 1.0 \times 10^{-4} \ kg/m$,we have:
$8.0 \times 10^{-3} = \frac{1}{2f \sqrt{1.0 \times 10^{-4}}} = \frac{1}{2f \times 10^{-2}}$.
Solving for $f$: $f = \frac{1}{2 \times 10^{-2} \times 8.0 \times 10^{-3}} = \frac{1}{16 \times 10^{-5}} = \frac{10^5}{16} = 6250 \ Hz$.
To find the error $\Delta f$,we use the relative error formula from $f = \frac{1}{2m\sqrt{\mu}}$:
$\frac{\Delta f}{f} = \frac{\Delta m}{m} + \frac{1}{2} \frac{\Delta \mu}{\mu}$.
Given $\Delta m = 0.3 \times 10^{-3}$ and $\Delta \mu = 0.1 \times 10^{-4}$:
$\frac{\Delta f}{f} = \frac{0.3 \times 10^{-3}}{8.0 \times 10^{-3}} + \frac{1}{2} \left( \frac{0.1 \times 10^{-4}}{1.0 \times 10^{-4}} \right) = \frac{0.3}{8.0} + \frac{0.1}{2} = 0.0375 + 0.05 = 0.0875$.
$\Delta f = 6250 \times 0.0875 = 546.875 \ Hz \approx 546.9 \ Hz$.
Thus,$f = (6250 \pm 546.9) \ Hz$.