Beats and Tuning fork Questions in English

(C) The frequency of a tuning fork is a characteristic property of the source and depends on its physical dimensions and material properties,not on the medium's temperature.
Since the speed of sound in a medium depends on temperature $(v \propto \sqrt{T})$,and the relationship between speed,frequency,and wavelength is given by $v = f \lambda$,if the temperature changes,the velocity $v$ changes.
Since the frequency $f$ remains constant,the wavelength $\lambda$ must change to satisfy the equation $\lambda = v/f$.
Therefore,the correct option is $(c)$.

(D) The frequency of the tuning fork is $n = 384 \text{ Hz}$.
The velocity of sound in air is $v = 352 \text{ m/s}$.
The time taken for one vibration is $T = \frac{1}{n} = \frac{1}{384} \text{ s}$.
The distance traveled by sound in one vibration is the wavelength $\lambda = v \times T = \frac{v}{n} = \frac{352}{384} \text{ m}$.
For $36$ vibrations, the total time taken is $t = 36 \times T = \frac{36}{384} \text{ s}$.
The total distance traveled by the sound is $d = v \times t = 352 \times \frac{36}{384} \text{ m}$.
$d = \frac{352 \times 36}{384} = 33 \text{ m}$.

(C) When two sound waves of nearly equal frequencies superpose at a point, they produce the phenomenon of beats.
The resultant intensity $I$ at any point is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi)$, where $\phi$ is the phase difference.
For the intensity to be minimum $(I_{\min} = 0)$, the condition is $I_1 = I_2$ and $\cos(\phi) = -1$.
This implies that the amplitudes of the two $S.H.M.$ produced by the sources must be equal, and they must be traveling along the same straight line to interfere destructively at the point of observation.

(C) Let the two tuning forks be $A$ and $B$ with frequencies $n_A = 256 \, Hz$ (known) and $n_B$ (unknown). The beat frequency is $x = 4 \, bps$.
The possible frequencies for the unknown tuning fork are $n_B = 256 + 4 = 260 \, Hz$ or $n_B = 256 - 4 = 252 \, Hz$.
It is given that when the fork $A$ (frequency $256 \, Hz$) is loaded with wax,its frequency decreases $(n_A \downarrow)$. Consequently,the beat frequency increases.
If $n_B = 252 \, Hz$,then $n_A - n_B = 256 - 252 = 4 \, Hz$. When $n_A$ decreases,the difference $(n_A - n_B)$ decreases,meaning the beat frequency would decrease.
If $n_B = 260 \, Hz$,then $n_B - n_A = 260 - 256 = 4 \, Hz$. When $n_A$ decreases,the difference $(n_B - n_A)$ increases,meaning the beat frequency increases.
Since the problem states that the beat frequency increases,the correct frequency of the other fork must be $n_B = 260 \, Hz$.

(D) Beats are the result of the superposition of two waves of nearly equal frequencies.
When two sound waves with slightly different frequencies travel in the same direction,they interfere with each other.
This interference causes the resultant amplitude to vary periodically with time,leading to a phenomenon known as beats.
The frequency of the beats is equal to the difference between the frequencies of the two original waves,i.e.,$f_{beat} = |f_1 - f_2|$.
As the waves alternately interfere constructively and destructively,the loudness of the sound rises and falls periodically.

(C) The phenomenon of beats occurs when two sound waves of slightly different frequencies interfere with each other.
The beat frequency is defined as the absolute difference between the frequencies of the two interfering sound waves.
If the frequencies of the two notes are $n_1$ and $n_2$,the number of beats heard per second is given by the formula:
$B = |n_1 - n_2|$
Therefore,the correct option is $C$.

(A) Let the known frequency be $n_A = 100 \, Hz$ and the unknown frequency be $n_B$.
The beat frequency is given by $|n_A - n_B| = 2 \, Hz$. This implies $n_B = 100 \pm 2$,so $n_B$ is either $102 \, Hz$ or $98 \, Hz$.
When the unknown tuning fork is loaded,its frequency $n_B$ decreases $(n_B \downarrow)$.
Case $1$: If $n_B = 98 \, Hz$,loading it further decreases the frequency (e.g.,to $97 \, Hz$). The new beat frequency would be $|100 - 97| = 3 \, Hz$,which is an increase.
Case $2$: If $n_B = 102 \, Hz$,loading it decreases the frequency (e.g.,to $101 \, Hz$). The new beat frequency would be $|100 - 101| = 1 \, Hz$,which is a decrease.
Since the problem states that the beat frequency decreases to $1 \, Hz$ after loading,the unknown frequency must be $102 \, Hz$.

(D) Let the frequency of the unknown tuning fork be $n_B$ and the known frequency be $n_A = 256 \ Hz$.
The beat frequency is given by $|n_A - n_B| = 2 \ Hz$.
This implies $n_B = 256 \pm 2$,so $n_B$ is either $258 \ Hz$ or $254 \ Hz$.
When the known tuning fork $(256 \ Hz)$ is loaded with wax,its frequency $n_A$ decreases.
Given that the beat frequency decreases from $2 \ Hz$ to $1 \ Hz$ after loading,we analyze the two cases:
Case $1$: If $n_B = 258 \ Hz$,then $n_B - n_A = 258 - 256 = 2 \ Hz$. After loading,$n_A$ decreases,so $(n_B - n_A)$ increases. This contradicts the observation.
Case $2$: If $n_B = 254 \ Hz$,then $n_A - n_B = 256 - 254 = 2 \ Hz$. After loading,$n_A$ decreases,so $(n_A - n_B)$ decreases. This matches the observation $(1 \ Hz)$.
Therefore,the frequency of the unknown tuning fork is $254 \ Hz$.

(B) Given: Frequency of tuning fork $A$ $(n_A)$ = $256 \ Hz$.
Initial beat frequency $(x)$ = $4 \ Hz$.
When $A$ is loaded with wax,its frequency decreases $(n_A \downarrow)$.
New beat frequency $(x')$ = $2 \ Hz$.
Since the beat frequency decreases from $4 \ Hz$ to $2 \ Hz$ upon loading $A$,the initial condition must be $n_A - n_B = x$.
Substituting the values: $256 - n_B = 4$,which gives $n_B = 252 \ Hz$.
Verification: After loading $A$,$n_A$ becomes less than $256$. Let the new frequency be $n_A'$. Then $n_A' - n_B = 2$. Since $n_B = 252$,$n_A' = 254 \ Hz$. This is consistent with the frequency of $A$ decreasing.
Therefore,the frequency of $B$ is $252 \ Hz$.

(C) The beat frequency is given by $f_b = |n_1 - n_2| = |260 - 256| = 4 \text{ Hz}$.
The time period of the beat is $T = \frac{1}{f_b} = \frac{1}{4} \text{ s}$.
At $t = 0$,the intensity is maximum,which corresponds to a phase of $0$ or $2\pi$.
The phase difference $\phi$ at time $t$ is related to the beat period $T$ by the formula $\phi = \frac{2\pi}{T} \times t$.
Substituting the given values: $\phi = \frac{2\pi}{(1/4)} \times \frac{1}{16} = 8\pi \times \frac{1}{16} = \frac{\pi}{2}$ radians.

(A) The number of beats produced per second is equal to the difference in frequencies of the two tuning forks.
Beat frequency $b = |n_2 - n_1| = |454\, Hz - 450\, Hz| = 4\, Hz$.
The time interval between two successive maximum intensities is the time period of the beats,which is the reciprocal of the beat frequency.
Time interval $T = \frac{1}{b} = \frac{1}{4}\, sec = 0.25\, sec$.

(D) Let the frequency of the first tuning fork be $n_A = 341 \ Hz$ and the frequency of the second tuning fork be $n_B$.
The beat frequency is given by $|n_A - n_B| = 6 \ Hz$.
This implies $n_B = 341 \pm 6$,so $n_B$ could be $347 \ Hz$ or $335 \ Hz$.
When the second tuning fork is loaded with wax,its frequency $n_B$ decreases $(n_B \downarrow)$.
After loading,the new beat frequency is $2 \ Hz$.
Case $1$: If $n_B = 335 \ Hz$,then $n_A - n_B = 6 \ Hz$. Loading $n_B$ makes it smaller,so $(n_A - n_B)$ increases. This contradicts the observation that the beat frequency decreased to $2 \ Hz$.
Case $2$: If $n_B = 347 \ Hz$,then $n_B - n_A = 6 \ Hz$. Loading $n_B$ makes it smaller,so $(n_B - n_A)$ decreases. As $n_B$ decreases towards $n_A$,the beat frequency decreases. This matches the observation that the beat frequency decreased to $2 \ Hz$.
Therefore,the natural frequency of the second tuning fork is $347 \ Hz$.

(B) The beat frequency $f_b$ is the difference between the frequencies of the two tuning forks.
$f_b = |f_2 - f_1| = |258 - 256| = 2 \text{ Hz}$.
The time interval between consecutive maxima (beats) is the reciprocal of the beat frequency.
$T = \frac{1}{f_b} = \frac{1}{2} = 0.5 \text{ sec}$.
Therefore,the time interval between consecutive maxima is $0.5 \text{ sec}$.

(C) Let the frequency of the known tuning fork be $n_A = 100\,Hz$ and the frequency of the unknown tuning fork be $n_B$.
Given that the beat frequency is $x = 5\,Hz$.
Therefore,$n_B = n_A \pm x$,which means $n_B = 100 \pm 5$,so $n_B$ is either $95\,Hz$ or $105\,Hz$.
When the tuning fork $B$ is loaded with wax,its frequency $n_B$ decreases.
If $n_B = 95\,Hz$,loading it further decreases the frequency,making the difference $|n_A - n_B|$ increase from $5\,Hz$ to a larger value.
If $n_B = 105\,Hz$,loading it decreases the frequency towards $100\,Hz$. The beat frequency $|n_B - n_A|$ decreases initially. However,if the loading is such that the frequency remains $105\,Hz$ (or if the question implies the beat frequency remains $5\,Hz$ after loading),we analyze the condition:
Since the beat frequency remains $5\,Hz$ after loading,it implies the frequency was $105\,Hz$ and decreased,but the beat frequency is defined by the absolute difference.
For the beat frequency to remain $5\,Hz$ after loading (decreasing $n_B$),$n_B$ must have been greater than $n_A$.
Thus,$n_B = 100 + 5 = 105\,Hz$.

(B) Let the frequency of tuning fork $F_1$ be $n_1 = 256 Hz$ and the frequency of tuning fork $F_2$ be $n_2$.
Given that the beat frequency is $6 Hz$,so $|n_1 - n_2| = 6$.
This implies $n_2 = 256 \pm 6$,so $n_2 = 262 Hz$ or $n_2 = 250 Hz$.
When $F_2$ is loaded with wax,its frequency decreases $(n_2' < n_2)$.
If $n_2 = 250 Hz$,loading it would decrease the frequency further away from $256 Hz$,increasing the beat frequency (e.g.,$256 - 249 = 7 Hz$),which contradicts the observation that the beat frequency remains $6 Hz$.
If $n_2 = 262 Hz$,loading it decreases the frequency towards $256 Hz$. The new beat frequency is $n_2' - n_1 = 6 Hz$,which is consistent with the problem statement.
Therefore,the initial frequency of $F_2$ was $262 Hz$.

(C) The beat frequency is given by $|n_A - n_B| = 5 \ Hz$. Since $n_B = 512 \ Hz$,$n_A$ can be either $517 \ Hz$ or $507 \ Hz$.
When a tuning fork is filed,its mass decreases,which causes its frequency to increase $(n_A \uparrow)$.
It is given that the beat frequency increases after filing.
Case $1$: If $n_A = 507 \ Hz$,then $n_B - n_A = 5 \ Hz$. As $n_A$ increases,the difference $(n_B - n_A)$ decreases (i.e.,beats decrease).
Case $2$: If $n_A = 517 \ Hz$,then $n_A - n_B = 5 \ Hz$. As $n_A$ increases,the difference $(n_A - n_B)$ increases (i.e.,beats increase).
Since the problem states that the number of beats increases,the correct frequency of $A$ must be $517 \ Hz$.

(C) The intensity of a sound wave is directly proportional to the square of its amplitude,i.e.,$I \propto A^2$.
When two sound sources of the same amplitude $a_0$ interfere,the maximum amplitude $A_{\max}$ produced during the formation of beats is the sum of the individual amplitudes: $A_{\max} = a_0 + a_0 = 2a_0$.
The maximum intensity $I_{\max}$ is then given by $I_{\max} \propto (A_{\max})^2 = (2a_0)^2 = 4a_0^2$.
Since the intensity of one source is $I_0 \propto a_0^2$,we have $I_{\max} = 4I_0$.
Therefore,the maximum intensity of beats is $4$ times that of one source.

(C) The number of beats heard per second is given by the difference in frequencies of the two waves,$|n_1 - n_2|$.
Given the equations $y_1 = a \sin(2000 \pi t)$ and $y_2 = a \sin(2008 \pi t)$,we compare these with the standard form $y = a \sin(2 \pi n t)$.
For the first wave: $2 \pi n_1 = 2000 \pi$,which gives $n_1 = 1000 \text{ Hz}$.
For the second wave: $2 \pi n_2 = 2008 \pi$,which gives $n_2 = 1004 \text{ Hz}$.
The number of beats per second is $|n_2 - n_1| = |1004 - 1000| = 4 \text{ beats per second}$.

(C) Let the actual frequency of the tuning fork be $f$.
When the oscillator frequency $f_1 = 514 \ Hz$ is used,the beat frequency is $|f - 514| = 2 \ Hz$. This implies $f = 514 \pm 2$,so $f = 516 \ Hz$ or $f = 512 \ Hz$.
When the oscillator frequency $f_2 = 510 \ Hz$ is used,the beat frequency is $|f - 510| = 6 \ Hz$. This implies $f = 510 \pm 6$,so $f = 516 \ Hz$ or $f = 504 \ Hz$.
The common frequency in both cases is $516 \ Hz$.
Thus,the actual frequency of the tuning fork is $516 \ Hz$.

(B) Let $n_S$ be the frequency of the sonometer string and $n_f = 480 \ Hz$ be the frequency of the tuning fork.
The beat frequency is given by $|n_f - n_S| = 10 \ Hz$.
This implies $n_S = 480 \pm 10$,so $n_S$ is either $470 \ Hz$ or $490 \ Hz$.
The frequency of a sonometer string is given by $n_S = \frac{1}{2l} \sqrt{\frac{T}{m}}$. Thus,$n_S \propto \sqrt{T}$.
When the tension $T$ is increased,$n_S$ increases.
Case $1$: If $n_S = 470 \ Hz$,then $n_f - n_S = 10 \ Hz$. As $n_S$ increases,$n_f - n_S$ decreases (e.g.,if $n_S$ becomes $475 \ Hz$,beats become $5 \ Hz$). This matches the given condition.
Case $2$: If $n_S = 490 \ Hz$,then $n_S - n_f = 10 \ Hz$. As $n_S$ increases,$n_S - n_f$ increases (e.g.,if $n_S$ becomes $495 \ Hz$,beats become $15 \ Hz$). This contradicts the given condition.
Therefore,the frequency of the string must be $470 \ Hz$.

(C) Let the frequency of tuning fork $A$ be $n_A$ and the frequency of tuning fork $B$ be $n_B = 256 \ Hz$.
The beat frequency is given as $x = 3 \ Hz$.
Therefore,$n_A = n_B \pm x = 256 \pm 3$,which means $n_A$ is either $259 \ Hz$ or $253 \ Hz$.
When tuning fork $A$ is loaded with wax,its frequency $n_A$ decreases.
Case $1$: If $n_A = 253 \ Hz$,loading it will decrease the frequency further below $253 \ Hz$. The new beat frequency would be $256 - n_A' > 3$,which contradicts the observation that the beat frequency remains $3 \ Hz$.
Case $2$: If $n_A = 259 \ Hz$,loading it will decrease the frequency towards $256 \ Hz$. The new beat frequency would be $n_A' - 256 = 3$,which is consistent with the observation.
Thus,the original frequency of tuning fork $A$ is $259 \ Hz$.

(A) Let the frequency of the source be $f$.
Since it produces $5$ beats per second with a source of $100 \, s^{-1}$,the frequency $f$ can be $100 \pm 5$,which is $105 \, s^{-1}$ or $95 \, s^{-1}$.
The second harmonic of the source is $2f$.
If $f = 105 \, s^{-1}$,the second harmonic is $210 \, s^{-1}$.
If $f = 95 \, s^{-1}$,the second harmonic is $190 \, s^{-1}$.
We are given that the second harmonic produces $5$ beats per second with a source of $205 \, s^{-1}$.
Checking the cases:
For $210 \, s^{-1}$: $|210 - 205| = 5 \, s^{-1}$ (This matches the condition).
For $190 \, s^{-1}$: $|190 - 205| = 15 \, s^{-1}$ (This does not match).
Therefore,the frequency of the source is $105 \, s^{-1}$.

(D) Beats are the periodic variation in the intensity of sound heard when two sound waves of slightly different frequencies,but similar amplitudes,are superimposed on each other.
If the frequencies of the two waves are $f_1$ and $f_2$,the beat frequency is given by $f_{beat} = |f_1 - f_2|$.
Therefore,the condition for producing beats is that the waves must have different frequencies.

(B) Initial frequency of $A$ is $n_A = 256 \ Hz$. The initial beat frequency is $x_1 = 4 \ Hz$.
Since $n_B = n_A \pm x_1$,the initial frequency of $B$ could be $256 + 4 = 260 \ Hz$ or $256 - 4 = 252 \ Hz$.
After loading $B$ slightly,its frequency $n_B$ decreases. The new beat frequency is $x_2 = 5 \text{ beats} / 2 \text{ seconds} = 2.5 \ Hz$.
Case $1$: If initial $n_B = 252 \ Hz$,loading $B$ decreases its frequency further away from $256 \ Hz$,so the beat frequency should increase. This contradicts the observation that the beat frequency decreased from $4 \ Hz$ to $2.5 \ Hz$.
Case $2$: If initial $n_B = 260 \ Hz$,loading $B$ decreases its frequency towards $256 \ Hz$,so the beat frequency should decrease. The new frequency is $n_B' = 256 + 2.5 = 258.5 \ Hz$.
However,the question asks for the frequency of $B$ after loading. Since $n_B' = 258.5 \ Hz$,the beat frequency is $258.5 - 256 = 2.5 \ Hz$,which matches the condition. Thus,the frequency of $B$ after loading is $258.5 \ Hz$.

(D) The frequency of tuning fork $A$ is $n_A = 200 \ Hz$. The beat frequency is $5 \ Hz$,so the frequency of fork $B$ $(n_B)$ is either $200 + 5 = 205 \ Hz$ or $200 - 5 = 195 \ Hz$.
When wax is added to fork $A$,its frequency $(n_A)$ decreases.
If $n_B = 195 \ Hz$,then the new beat frequency would be $n_A' - n_B$ (where $n_A' < 200$). As $n_A$ decreases,the difference $n_A - n_B$ decreases,so the beat frequency would decrease from $5 \ Hz$ to a lower value.
If $n_B = 205 \ Hz$,then the beat frequency is $n_B - n_A = 205 - 200 = 5 \ Hz$. As $n_A$ decreases,the difference $n_B - n_A$ increases,so the beat frequency increases from $5 \ Hz$ to $8 \ Hz$.
Since the beat frequency increases to $8 \ Hz$,the correct frequency of fork $B$ must be $205 \ Hz$.

(C) The frequency of tuning fork $A$ is $n_A = 320 \ Hz$. The beat frequency is $4 \ Hz$,so the frequency of $B$ $(n_B)$ is either $320 + 4 = 324 \ Hz$ or $320 - 4 = 316 \ Hz$.
When wax is added to a tuning fork,its frequency decreases.
If $n_B = 316 \ Hz$,adding wax would decrease the frequency further,increasing the beat frequency $(320 - n_B' > 4)$,which contradicts the problem statement.
If $n_B = 324 \ Hz$,adding wax decreases the frequency of $B$ $(n_B' < 324)$. For the beat frequency to remain $4 \ Hz$,the new frequency must be $n_B' = 320 + 4 = 324 \ Hz$ (impossible as it decreased) or $n_B' = 320 - 4 = 316 \ Hz$ (also impossible). Wait,let's re-evaluate: If $n_B = 316 \ Hz$,adding wax makes $n_B' < 316$,so $|320 - n_B'| > 4$. If $n_B = 324 \ Hz$,adding wax makes $n_B' < 324$. If $n_B'$ becomes $316 \ Hz$,the beats would change. However,the standard interpretation for this classic problem is that $n_B = 324 \ Hz$ is the correct value because adding wax to $324 \ Hz$ brings it closer to $320 \ Hz$,but the question states $4$ beats are heard again. This implies $n_B - n_A = 4$ initially $(324 - 320 = 4)$. After adding wax,$n_B$ decreases to $n_B'$. If $n_B' - n_A = 4$ is not possible,we check $n_A - n_B' = 4$. If $n_B = 316 \ Hz$,$n_A - n_B = 4$. Adding wax makes $n_B' < 316$,so $n_A - n_B' > 4$. Thus,the initial frequency must have been $n_B = 324 \ Hz$.

(C) The beat frequency is given by $f_b = |f_1 - f_2| = |384 - 380| = 4 \ Hz$.
The time period of one complete beat cycle is $T = \frac{1}{f_b} = \frac{1}{4} = 0.25 \ \text{seconds}$.
$A$ beat cycle consists of one maximum (loud sound) and one minimum (faint sound).
The time interval between a maximum sound and the subsequent minimum sound is half of the beat period.
Therefore,$t = \frac{T}{2} = \frac{0.25}{2} = 0.125 \ \text{seconds}$.

(D) The intensity $I$ of a sound wave is proportional to the square of its amplitude $a$,i.e.,$I \propto a^2$.
For two waves with amplitudes $a_1 = 5$ and $a_2 = 3$,the maximum intensity $I_{\max}$ occurs when the waves interfere constructively,and the minimum intensity $I_{\min}$ occurs when they interfere destructively.
The ratio of maximum to minimum intensity is given by the formula:
$\frac{I_{\max}}{I_{\min}} = \left( \frac{a_1 + a_2}{a_1 - a_2} \right)^2$
Substituting the given values:
$\frac{I_{\max}}{I_{\min}} = \left( \frac{5 + 3}{5 - 3} \right)^2 = \left( \frac{8}{2} \right)^2 = (4)^2 = \frac{16}{1}$
Thus,the ratio is $16:1$.

(A) The frequency of a wave is given by $n = \frac{v}{\lambda}$,where $v$ is the velocity and $\lambda$ is the wavelength.
Given wavelengths are $\lambda_1 = 50 \; cm = 0.50 \; m$ and $\lambda_2 = 51 \; cm = 0.51 \; m$.
The frequencies are $n_1 = \frac{v}{0.50}$ and $n_2 = \frac{v}{0.51}$.
The beat frequency is the difference between the two frequencies: $\Delta n = n_1 - n_2 = 12 \; Hz$.
Substituting the values: $v \left( \frac{1}{0.50} - \frac{1}{0.51} \right) = 12$.
$v \left( \frac{0.51 - 0.50}{0.50 \times 0.51} \right) = 12$.
$v \left( \frac{0.01}{0.255} \right) = 12$.
$v = \frac{12 \times 0.255}{0.01} = 12 \times 25.5 = 306 \; m/s$.

(C) The general equation for a wave is $y = A \sin(\omega t)$.
Comparing the given equations with the standard form,we have angular frequencies $\omega_1 = 316 \text{ rad/s}$ and $\omega_2 = 310 \text{ rad/s}$.
The frequency $f$ is related to angular frequency $\omega$ by the formula $f = \frac{\omega}{2\pi}$.
Therefore,the frequencies of the two waves are $f_1 = \frac{316}{2\pi} \text{ Hz}$ and $f_2 = \frac{310}{2\pi} \text{ Hz}$.
The number of beats produced per second is the difference between the two frequencies:
$\text{Beat frequency} = |f_1 - f_2| = \left| \frac{316}{2\pi} - \frac{310}{2\pi} \right| = \frac{6}{2\pi} = \frac{3}{\pi} \text{ Hz}$.

(B) The beat frequency $(f_b)$ is defined as the number of beats produced per unit time.
Given:
Number of beats $(n)$ = $2$
Time interval $(t)$ = $0.4 \ s$
Formula: $f_b = \frac{n}{t}$
Calculation: $f_b = \frac{2}{0.4} = \frac{20}{4} = 5 \ Hz$.
Therefore,the beat frequency is $5 \ Hz$.

(A) The number of beats per second is the difference between the two frequencies.
Given that the unknown frequency $x$ produces $8$ beats per second with $250 \ Hz$,the possible values for $x$ are $250 \pm 8$,which are $258 \ Hz$ or $242 \ Hz$.
Given that the unknown frequency $x$ produces $12$ beats per second with $270 \ Hz$,the possible values for $x$ are $270 \pm 12$,which are $282 \ Hz$ or $258 \ Hz$.
Comparing both sets of possibilities,the common frequency is $258 \ Hz$.
Therefore,the unknown frequency $x$ is $258 \ Hz$.

(A) The speed of sound at $0^{\circ}C$ is $v_0 = 332 \ m/s$.
The speed of sound at $t^{\circ}C$ is given by $v_t = v_0 + 0.61 \times t$.
At $t = 20^{\circ}C$,the speed of sound is $v_{20} = 332 + 0.61 \times 20 = 332 + 12.2 = 344.2 \ m/s$.
The frequencies of the two waves are $n_1 = \frac{v}{\lambda_1}$ and $n_2 = \frac{v}{\lambda_2}$.
The number of beats produced per second is the difference in frequencies: $\Delta n = |n_1 - n_2| = v \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right)$.
Converting wavelengths to meters: $\lambda_1 = 0.50 \ m$ and $\lambda_2 = 0.51 \ m$.
$\Delta n = 344.2 \left( \frac{1}{0.50} - \frac{1}{0.51} \right) = 344.2 \left( 2 - 1.96078 \right) = 344.2 \times 0.03922 \approx 13.5 \approx 14 \ Hz$.

(A) The human ear has a phenomenon known as the persistence of hearing,which lasts for approximately $0.1 \ s$.
This means that if two sounds reach the ear with a time interval of less than $0.1 \ s$,the ear cannot distinguish them as separate sounds.
Therefore,the maximum number of beats per second that can be heard is given by the reciprocal of the persistence of hearing:
$f_{max} = \frac{1}{0.1 \ s} = 10 \ Hz$.
Thus,the human ear can distinguish a maximum of $10$ beats per second.

(D) Let the frequency of tuning fork $A$ be $n_A$ and the frequency of tuning fork $B$ be $n_B = 384 \ Hz$.
The initial beat frequency is $x_1 = |n_A - n_B| = 6 \ Hz$.
This implies $n_A = 384 \pm 6$,so $n_A$ is either $390 \ Hz$ or $378 \ Hz$.
When tuning fork $A$ is loaded with wax,its frequency $n_A$ decreases.
After loading,the new beat frequency is $x_2 = 4 \ Hz$.
If $n_A = 390 \ Hz$,loading decreases it towards $384 \ Hz$,so the beat frequency decreases from $6$ to $4$. This matches the given condition.
If $n_A = 378 \ Hz$,loading decreases it further away from $384 \ Hz$,so the beat frequency would increase from $6$ to a higher value. This contradicts the given condition.
Therefore,the original frequency of tuning fork $A$ is $390 \ Hz$.

(B) The phenomenon of beats occurs when two sound waves of slightly different frequencies interfere with each other.
For the human ear to perceive these beats,the frequency difference between the two sources must be small,typically less than $10 \ Hz$ to $15 \ Hz$,due to the persistence of hearing.
In option $A$,the difference is $150 \ Hz - 100 \ Hz = 50 \ Hz$.
In option $B$,the difference is $25 \ Hz - 20 \ Hz = 5 \ Hz$.
In option $C$,the difference is $500 \ Hz - 400 \ Hz = 100 \ Hz$.
In option $D$,the difference is $1500 \ Hz - 1000 \ Hz = 500 \ Hz$.
Since only the frequency difference in option $B$ is within the audible range for beats,it is the correct answer.

(A) The number of beats per second (beat frequency) is given by $n = \frac{30}{3} = 10 \ Hz$.
The beat frequency is also given by the formula $n = v \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right)$,where $v$ is the velocity of sound.
Substituting the given values: $10 = v \left( \frac{1}{5} - \frac{1}{6} \right)$.
Calculating the difference: $\frac{1}{5} - \frac{1}{6} = \frac{6-5}{30} = \frac{1}{30}$.
Therefore,$10 = v \left( \frac{1}{30} \right)$.
Solving for $v$: $v = 10 \times 30 = 300 \ m/s$.

(A) The frequency of a wave is given by $n = \frac{v}{\lambda}$,where $v$ is the speed of sound and $\lambda$ is the wavelength.
Given: $v = 396 \ m/s$,$\lambda_1 = 99 \ cm = 0.99 \ m$,and $\lambda_2 = 100 \ cm = 1.0 \ m$.
The frequencies are $n_1 = \frac{396}{0.99} = 400 \ Hz$ and $n_2 = \frac{396}{1.0} = 396 \ Hz$.
The number of beats heard per second is the difference in frequencies: $\Delta n = |n_1 - n_2| = |400 - 396| = 4 \ Hz$.
Therefore,the number of beats heard is $4$.

(D) Let the frequency of the known tuning fork be $n_A = 288 \, Hz$ and the unknown fork be $n_B$.
Initially,the beat frequency is $x = |n_A - n_B| = 4 \, Hz$.
This implies $n_B = 288 \pm 4$,so $n_B$ is either $292 \, Hz$ or $284 \, Hz$.
When wax is added to the unknown fork,its frequency $n_B$ decreases $(n_B \downarrow)$.
After adding wax,the new beat frequency is $x' = 2 \, Hz$.
Case $1$: If $n_B = 284 \, Hz$,then $n_B$ decreases further away from $288 \, Hz$,so the beat frequency would increase $(|288 - 283| = 5 \, Hz)$,which contradicts the observation.
Case $2$: If $n_B = 292 \, Hz$,then $n_B$ decreases towards $288 \, Hz$,so the beat frequency decreases $(|288 - 290| = 2 \, Hz)$.
This matches the given observation.
Therefore,the frequency of the unknown fork is $292 \, Hz$.

(A) The frequency $(f)$ of a vibrating source is defined as the number of oscillations or cycles per unit time.
Given that the tuning fork produces $2$ beats in $0.04$ seconds.
Using the formula: $f = \frac{\text{Number of beats}}{\text{Time in seconds}}$
$f = \frac{2}{0.04} = \frac{200}{4} = 50\,Hz$.
Therefore,the frequency of the tuning fork is $50\,Hz$.

(C) The number of beats produced per unit time is defined as the beat frequency.
Beat frequency is equal to the difference between the frequencies of the two sound sources.
Given that $4$ beats are produced in $0.25 \ s$.
Beat frequency = $\frac{\text{Number of beats}}{\text{Time interval}} = \frac{4}{0.25} \ Hz$.
Beat frequency = $16 \ Hz$.
Therefore,the difference in their frequencies is $16 \ Hz$.

(D) Let the frequency of the piano string be $n_P$. The frequency of the tuning fork is $n_f = 256\,Hz$.
The beat frequency is given by $|n_f - n_P| = 5\,Hz$. This implies $n_P = 256 \pm 5\,Hz$,so $n_P$ is either $251\,Hz$ or $261\,Hz$.
When the tension $T$ in the piano string is increased,the frequency $n_P$ increases because $n_P \propto \sqrt{T}$.
If $n_P$ was $251\,Hz$,increasing the tension makes $n_P$ approach $256\,Hz$,which would decrease the beat frequency $|256 - n_P|$.
If $n_P$ was $261\,Hz$,increasing the tension makes $n_P$ move further away from $256\,Hz$,which would increase the beat frequency.
Since the beat frequency decreases from $5\,Hz$ to $2\,Hz$,the initial frequency must have been $251\,Hz$.
Therefore,the frequency of the piano string before increasing the tension was $256 - 5 = 251\,Hz$.

(B) The frequency of a tuning fork is given by the relation $n = \frac{1}{2l} \sqrt{\frac{Y}{\rho}}$,where $Y$ is Young's modulus and $\rho$ is the density of the material.
As the temperature increases,the Young's modulus $Y$ of the material decreases,and the length $l$ of the prongs increases due to thermal expansion.
Both these factors contribute to a decrease in the frequency of the tuning fork.
The relationship is given by $n_t = n_0(1 - \alpha t)$,where $n_t$ is the frequency at $t^\circ C$,$n_0$ is the frequency at $0^\circ C$,and $\alpha$ is a constant related to the material properties.
Therefore,the correct option is $B$.

(A) Given: Frequency of string $X$ $(n_X)$ = $300 \ Hz$.
Initial beat frequency = $|n_X - n_Y| = 4 \ Hz$.
This implies the initial frequency of $Y$ $(n_Y)$ is either $300 - 4 = 296 \ Hz$ or $300 + 4 = 304 \ Hz$.
When the tension of string $Y$ is increased,its frequency $n_Y$ increases because $n \propto \sqrt{T}$.
Case $1$: If $n_Y = 296 \ Hz$,increasing the tension makes $n_Y$ approach $300 \ Hz$ (e.g.,$298 \ Hz$),so the beat frequency $|300 - 298| = 2 \ Hz$. This matches the given condition.
Case $2$: If $n_Y = 304 \ Hz$,increasing the tension makes $n_Y$ move further away from $300 \ Hz$ (e.g.,$306 \ Hz$),so the beat frequency $|300 - 306| = 6 \ Hz$. This does not match the condition.
Therefore,the original frequency of $Y$ was $296 \ Hz$.

(C) Let $n$ be the frequency of tuning fork $C$ in $Hz$.
Given that the frequency of $A$ is $3\%$ more than $C$,so $n_A = n + 0.03n = 1.03n$.
The frequency of $B$ is $2\%$ less than $C$,so $n_B = n - 0.02n = 0.98n$.
When $A$ and $B$ are sounded together,the beat frequency is given by $|n_A - n_B| = 5$.
Substituting the expressions: $|1.03n - 0.98n| = 5$.
$0.05n = 5$.
$n = \frac{5}{0.05} = 100\,Hz$.
Therefore,the frequency of tuning fork $A$ is $n_A = 1.03 \times 100 = 103\,Hz$.

(B) The given equations for the progressive waves are $Y_1 = 4\sin(500\pi t)$ and $Y_2 = 2\sin(506\pi t)$.
Comparing these with the standard equation $Y = A\sin(\omega t)$,we get angular frequencies $\omega_1 = 500\pi \text{ rad/s}$ and $\omega_2 = 506\pi \text{ rad/s}$.
The frequencies $n_1$ and $n_2$ are calculated using $\omega = 2\pi n$:
$n_1 = \frac{500\pi}{2\pi} = 250 \text{ Hz}$
$n_2 = \frac{506\pi}{2\pi} = 253 \text{ Hz}$
The beat frequency is the difference between the two frequencies: $f_b = n_2 - n_1 = 253 - 250 = 3 \text{ beats per second}$.
To find the number of beats per minute,we multiply the beats per second by $60$:
$\text{Beats per minute} = 3 \times 60 = 180$.

(B) The frequency of a wave is determined by the source of the disturbance.
Since the tuning fork acts as the source,the frequency of the sound wave produced remains constant regardless of the medium through which it travels.
Therefore,the frequency is the same in both the material of the tuning fork and the air.
Other properties like wavelength and velocity depend on the medium's properties (such as density and elasticity),and amplitude depends on the energy of the wave,which may change due to damping.

(B) The frequency $f$ of the sound produced by a siren disc is given by the formula $f = n \times \nu$,where $n$ is the number of holes and $\nu$ is the frequency of rotation in revolutions per second.
Given,$n = 60$ holes.
The rotational speed is $360\,rpm$ (revolutions per minute).
To convert $rpm$ to revolutions per second,divide by $60$: $\nu = \frac{360}{60} = 6\,rev/s$.
Therefore,the frequency of the emitted sound is $f = 60 \times 6 = 360\,Hz$.
Since the sound is in unison with the tuning fork,the frequency of the tuning fork is $360\,Hz$.

(C) The frequency of a stretched wire is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,which implies $n \propto \sqrt{T}$.
Taking the derivative,we get $\frac{\Delta n}{n} = \frac{1}{2} \frac{\Delta T}{T}$.
Given that the tension increases by $2\%$,i.e.,$\frac{\Delta T}{T} = 0.02$,the change in frequency is $\frac{\Delta n}{n} = \frac{1}{2} \times 0.02 = 0.01$ or $1\%$.
Let the initial frequency be $n$. The new frequency is $n' = n + 0.01n = 1.01n$.
The number of beats produced per second is the difference in frequencies: $n' - n = 5$.
$1.01n - n = 5 \implies 0.01n = 5$.
$n = \frac{5}{0.01} = 500 \ Hz$.

(D) Let the initial frequency of each wire be $n$. The frequency of a stretched wire is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,which implies $n \propto \sqrt{T}$.
Taking the derivative,we get $\frac{dn}{n} = \frac{1}{2} \frac{dT}{T}$.
Given the percentage change in tension is $1\%$,so $\frac{dT}{T} = 0.01$.
The change in frequency is $\Delta n = \frac{1}{2} \times n \times \frac{dT}{T} = \frac{1}{2} \times n \times 0.01 = 0.005n$.
We are given that $3$ beats are heard in $2 \,s$,so the beat frequency is $\Delta n = \frac{3}{2} = 1.5 \,Hz$.
Equating the two expressions for $\Delta n$: $0.005n = 1.5$.
Solving for $n$: $n = \frac{1.5}{0.005} = 300 \,Hz$ or $300 \,s^{-1}$.