Beats and Tuning fork Questions in English

(A) Given: Frequency $f = 220 \,Hz$, Velocity $v = 330 \,m/s$.
First, calculate the wavelength $\lambda$ using the formula $v = f \lambda$:
$\lambda = \frac{v}{f} = \frac{330}{220} = 1.5 \,m$.
In one vibration, the sound wave travels a distance equal to one wavelength $\lambda$.
Therefore, for $80$ vibrations, the total distance $d$ is:
$d = 80 \times \lambda = 80 \times 1.5 = 120 \,m$.

(B) For a sonometer wire,the frequency $f$ is inversely proportional to the vibrating length $l$,i.e.,$f \propto 1/l$.
Let the frequencies of the two tuning forks be $f_1$ and $f_2$.
Given that $f_1 \propto 1/0.70$ and $f_2 \propto 1/0.69$.
Since $f_2 > f_1$,we have $f_2 - f_1 = 6 \ Hz$.
Let $f_1 = k/0.70$ and $f_2 = k/0.69$,where $k$ is a constant.
Substituting these into the beat frequency equation: $k/0.69 - k/0.70 = 6$.
$k(0.70 - 0.69) / (0.69 \times 0.70) = 6$.
$k(0.01) / 0.483 = 6$.
$k = 6 \times 0.483 / 0.01 = 6 \times 48.3 = 289.8$.
Now,$f_1 = 289.8 / 0.70 = 414 \ Hz$.
And $f_2 = 289.8 / 0.69 = 420 \ Hz$.
Thus,the frequencies are $414 \ Hz$ and $420 \ Hz$.

(C) Resonance occurs when the frequency of the driving force matches the natural frequency of the system or one of its harmonics (integer multiples).
Given frequency $f = 256 \ Hz$.
Harmonics are $n \times f$,where $n = 1, 2, 3, \dots$
$1st \ harmonic = 1 \times 256 = 256 \ Hz$.
$2nd \ harmonic = 2 \times 256 = 512 \ Hz$.
$3rd \ harmonic = 3 \times 256 = 768 \ Hz$.
Comparing the options:
$A) 256 \ Hz$ (Resonates)
$B) 512 \ Hz$ (Resonates)
$C) 754 \ Hz$ (Does not resonate)
$D) 768 \ Hz$ (Resonates)
Therefore,the tuning fork will not resonate with $754 \ Hz$.

(C) The given equations are $y_1 = 4 \sin(710 \pi t)$ and $y_2 = 3 \sin(702 \pi t)$.
Comparing these with $y = A \sin(2 \pi f t)$,we get:
$2 \pi f_1 = 710 \pi \implies f_1 = 355 \text{ Hz}$
$2 \pi f_2 = 702 \pi \implies f_2 = 351 \text{ Hz}$
The number of beats per second is $n = |f_1 - f_2| = |355 - 351| = 4 \text{ beats/s}$.
The amplitudes are $A_1 = 4$ and $A_2 = 3$.
The maximum intensity (waxing) is proportional to $(A_1 + A_2)^2 = (4 + 3)^2 = 7^2 = 49$.
The minimum intensity (waning) is proportional to $(A_1 - A_2)^2 = (4 - 3)^2 = 1^2 = 1$.
Therefore,the intensity ratio is $49:1$.
The correct answer is $4, 49:1$.

(B) The beat frequency is given by $|f_X - f_Y| = 6 \ Hz$.
Given $f_X = 300 \ Hz$,the possible frequencies for $f_Y$ are $300 - 6 = 294 \ Hz$ or $300 + 6 = 306 \ Hz$.
When the tension of a string is increased,its frequency $f$ increases because $f \propto \sqrt{T}$.
Case $1$: If $f_Y = 294 \ Hz$,increasing the tension increases $f_Y$. The new beat frequency would be $|300 - (294 + \Delta f)|$. Since the beat frequency decreases to $4 \ Hz$,the value must be approaching $300 \ Hz$. Thus,$300 - (294 + \Delta f) = 4$,which gives $\Delta f = 2 \ Hz$. This is possible.
Case $2$: If $f_Y = 306 \ Hz$,increasing the tension increases $f_Y$ further away from $300 \ Hz$. The new beat frequency would be $|300 - (306 + \Delta f)| = 6 + \Delta f$,which would be greater than $6 \ Hz$.
Since the beat frequency decreased,the original frequency must be $294 \ Hz$.

(D) The frequency of a vibrating wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$. Since the wires are identical,$L$ and $\mu$ are constant,so $f \propto \sqrt{T}$.
Given $T_1 > T_2$,the frequencies are $f_1$ and $f_2$ such that $f_1 > f_2$. The beat frequency is $f_1 - f_2 = 6$.
If we change the tension slightly,the beat frequency remains $6$ if the difference between the frequencies remains the same.
Let $f_1 = k\sqrt{T_1}$ and $f_2 = k\sqrt{T_2}$.
If $T_1$ is decreased,$f_1$ decreases,moving closer to $f_2$. If $T_2$ is increased,$f_2$ increases,moving closer to $f_1$. In both cases,the difference $f_1 - f_2$ decreases.
However,if we decrease $T_1$ such that $f_1$ becomes less than $f_2$,the beat frequency $|f_2 - f_1|$ can remain $6$ if the new difference is equal to the original difference.
Specifically,if $T_1$ is decreased or $T_2$ is increased,the frequencies approach each other and then cross. The condition for the beat frequency to remain unchanged after a slight change is that the new difference $|f_1' - f_2'|$ equals the original difference $6$.

(B) The frequency of a wave is given by $\omega = 2 \pi f$,where $\omega$ is the angular frequency.
For the first wave,$\omega_1 = 50 \pi$,so $f_1 = \frac{50 \pi}{2 \pi} = 25 \text{ Hz}$.
For the second wave,$\omega_2 = 46 \pi$,so $f_2 = \frac{46 \pi}{2 \pi} = 23 \text{ Hz}$.
The beat frequency is the difference between the two frequencies: $f_b = |f_1 - f_2| = |25 - 23| = 2 \text{ Hz}$.
The beat frequency represents the number of times the intensity reaches a maximum per second.
Therefore,the listener hears the sound of maximum intensity $2$ times in one second.

(D) The beat frequency $f_b$ is the difference between the frequencies of the two tuning forks.
$f_b = |f_1 - f_2| = |258 \ Hz - 256 \ Hz| = 2 \ Hz$.
The beat frequency represents the number of maxima (beats) produced per second.
The time interval between two consecutive maxima is the time period of the beats,denoted by $T_b$.
$T_b = \frac{1}{f_b} = \frac{1}{2 \ Hz} = 0.5 \ s$.
Therefore,the time interval between two consecutive maxima is $0.5 \ s$.

(B) Let $n$ be the frequency of the tuning fork.
Let $n_1$ and $n_2$ be the frequencies of the wire at tensions $T_1 = 225 \ N$ and $T_2 = 256 \ N$ respectively.
Since the frequency of a stretched wire is proportional to the square root of tension $(n \propto \sqrt{T})$,we have:
$n_1 = n - 6$ (or $n + 6$)
$n_2 = n + 6$ (or $n - 6$)
Since $T_2 > T_1$,it follows that $n_2 > n_1$. Thus,$n_1 = n - 6$ and $n_2 = n + 6$.
Taking the ratio:
$\frac{n_1}{n_2} = \sqrt{\frac{T_1}{T_2}} = \sqrt{\frac{225}{256}} = \frac{15}{16}$
$\frac{n - 6}{n + 6} = \frac{15}{16}$
$16(n - 6) = 15(n + 6)$
$16n - 96 = 15n + 90$
$n = 186 \ Hz$

(B) The general equation for a wave is given by $Y = A \sin(\omega t)$.
For the first wave,$Y_1 = 0.25 \sin(316t)$,comparing with the general equation,we get angular frequency $\omega_1 = 316 \text{ rad/s}$.
Since $\omega = 2\pi f$,the frequency $f_1 = \frac{\omega_1}{2\pi} = \frac{316}{2\pi} \text{ Hz}$.
For the second wave,$Y_2 = 0.25 \sin(310t)$,comparing with the general equation,we get angular frequency $\omega_2 = 310 \text{ rad/s}$.
The frequency $f_2 = \frac{\omega_2}{2\pi} = \frac{310}{2\pi} \text{ Hz}$.
The number of beats produced per second is the difference in frequencies: $f_{beat} = |f_1 - f_2|$.
$f_{beat} = \frac{316}{2\pi} - \frac{310}{2\pi} = \frac{6}{2\pi} = \frac{3}{\pi} \text{ Hz}$.

(B) Let the frequencies of the $28$ tuning forks be $n_1, n_2, \dots, n_{28}$ in increasing order.
Since each fork produces '$x$' beats per second with the preceding one,the frequencies form an Arithmetic Progression $(AP)$ with common difference $d = x$.
Thus,the frequency of the $k^{\text{th}}$ fork is $n_k = n_1 + (k-1)x$.
For the $12^{\text{th}}$ fork: $n_{12} = n_1 + 11x = 152 \text{ Hz} \quad \dots(1)$.
For the $28^{\text{th}}$ fork: $n_{28} = n_1 + 27x$.
Given that the last fork is an octave of the first,$n_{28} = 2n_1$.
Substituting the expression for $n_{28}$: $2n_1 = n_1 + 27x \Rightarrow n_1 = 27x$.
Substitute $n_1 = 27x$ into equation $(1)$:
$27x + 11x = 152$
$38x = 152$
$x = \frac{152}{38} = 4 \text{ Hz}$.

(C) Given equations of displacement are:
$x_1 = 2 \sin(1000 \pi t)$
$x_2 = 3 \sin(1006 \pi t)$
Comparing these with the standard equation $x = A \sin(\omega t)$,we get:
Angular frequencies: $\omega_1 = 1000 \pi$ and $\omega_2 = 1006 \pi$.
Amplitudes: $A_1 = 2$ and $A_2 = 3$.
The frequencies are $f_1 = \frac{\omega_1}{2 \pi} = \frac{1000 \pi}{2 \pi} = 500 \text{ Hz}$ and $f_2 = \frac{\omega_2}{2 \pi} = \frac{1006 \pi}{2 \pi} = 503 \text{ Hz}$.
Beat frequency is given by $|f_2 - f_1| = |503 - 500| = 3 \text{ beats/s}$.
Maximum intensity is proportional to the square of the maximum amplitude $(A_{\text{max}} = A_1 + A_2)$.
$A_{\text{max}} = 2 + 3 = 5$ units.
Maximum intensity $\propto (A_{\text{max}})^2 = (5)^2 = 25$ units.
Therefore,the waves produce $3$ beats/s with a maximum intensity of $25$ units.

(A) Let $n$ be the frequency of tuning fork $C$.
Given that the frequency of $A$ is $1.4 \%$ more than $C$, so $n_A = n + 0.014n = 1.014n$.
Given that the frequency of $B$ is $2.6 \%$ less than $C$, so $n_B = n - 0.026n = 0.974n$.
When $A$ and $B$ are sounded together, the number of beats produced per second is the difference in their frequencies: $|n_A - n_B| = 10$.
Substituting the values: $1.014n - 0.974n = 10$.
$0.04n = 10$.
$n = \frac{10}{0.04} = \frac{1000}{4} = 250 \text{ Hz}$.

(B) The beat frequency is defined as the difference between the frequencies of the two waves: $f_{beat} = |f_2 - f_1|$.
Given $f_1 = 340 \ Hz$ and $f_2 = 335 \ Hz$.
$f_{beat} = 340 \ Hz - 335 \ Hz = 5 \ Hz$.
The time interval between two successive maxima is the reciprocal of the beat frequency:
$T = \frac{1}{f_{beat}} = \frac{1}{5} \ s = 0.2 \ s$.

(D) The beat frequency is given by $f_b = |f_2 - f_1| = |256 \,Hz - 250 \,Hz| = 6 \,Hz$.
The time period of the beat wave is $T_b = \frac{1}{f_b} = \frac{1}{6} \,s$.
The intensity is maximum at $t=0$. The intensity becomes minimum at half the time period of the beat, which corresponds to the interval between a maximum and the subsequent minimum.
Therefore, the time required is $t = \frac{T_b}{2} = \frac{1}{6 \times 2} = \frac{1}{12} \,s$.

(C) The frequency of a vibrating wire is given by $f = \frac{1}{2l} \sqrt{\frac{T}{m}}$.
Since $T$ and $m$ are constant,$f \propto \frac{1}{l}$,which implies $fl = \text{constant}$.
Let $f$ be the frequency of the tuning fork.
Initially,the length $l_1 = 50 \ cm$ and the beat frequency is $3$,so the wire frequency $f_1 = f - 3$ (or $f + 3$).
After shortening the length by $1 \ cm$,$l_2 = 49 \ cm$. The beat frequency remains $3$,so the wire frequency $f_2 = f + 3$ (or $f - 3$).
Since $f_1 l_1 = f_2 l_2$,we have $(f - 3) \times 50 = (f + 3) \times 49$.
$50f - 150 = 49f + 147$.
$50f - 49f = 147 + 150$.
$f = 297 \ Hz$.

(A) The general equation for a wave is $y = A \sin(\omega t)$,where $\omega = 2 \pi f$.
Given $y_1 = 0.35 \sin(316 t)$,the angular frequency is $\omega_1 = 316 \text{ rad/s}$. Thus,the frequency $f_1 = \frac{\omega_1}{2 \pi} = \frac{316}{2 \pi} \text{ Hz}$.
Given $y_2 = 0.35 \sin(310 t)$,the angular frequency is $\omega_2 = 310 \text{ rad/s}$. Thus,the frequency $f_2 = \frac{\omega_2}{2 \pi} = \frac{310}{2 \pi} \text{ Hz}$.
The number of beats produced per second is the beat frequency $f_b = |f_1 - f_2|$.
$f_b = \frac{316}{2 \pi} - \frac{310}{2 \pi} = \frac{6}{2 \pi} = \frac{3}{\pi} \text{ beats per second}$.

(A) The general equation for a wave is given by $y = a \sin(2 \pi n t)$,where $n$ is the frequency.
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 beat frequency is the difference between the two frequencies: $f_{\text{beat}} = |n_2 - n_1| = |1004 - 1000| = 4 \text{ Hz}$.
Therefore,the number of beats heard per second is $4$.

(C) Given the frequencies of the tuning forks are $n_A, n_B, n_C$ such that $n_A > n_B > n_C$.
When forks $A$ and $B$ are sounded together,the beat frequency is $n_1 = n_A - n_B$ (Equation $i$).
When forks $A$ and $C$ are sounded together,the beat frequency is $n_2 = n_A - n_C$ (Equation $ii$).
We want to find the beat frequency when $B$ and $C$ are sounded together,which is $n_B - n_C$.
Subtracting Equation $i$ from Equation $ii$:
$(n_A - n_C) - (n_A - n_B) = n_2 - n_1$
$n_A - n_C - n_A + n_B = n_2 - n_1$
$n_B - n_C = n_2 - n_1$.
Thus,the number of beats produced per second is $n_2 - n_1$.

(A) Given: Wavelengths $\lambda_1 = 5.0 \ m$ and $\lambda_2 = 5.5 \ m$. Velocity of sound $v = 300 \ m/s$.
The frequency $n$ of a wave is given by $n = \frac{v}{\lambda}$.
Frequency of the first wave: $n_1 = \frac{300}{5.0} = 60 \ Hz$.
Frequency of the second wave: $n_2 = \frac{300}{5.5} = \frac{3000}{55} \approx 54.55 \ Hz$.
The number of beats produced per second is the difference in frequencies: $n_{beats} = |n_1 - n_2| = |60 - 54.55| = 5.45 \ Hz$.
Rounding to the nearest integer,the number of beats is approximately $5 \ Hz$ or $6 \ Hz$ depending on the precision. Given the options,$6 \ Hz$ is the closest match.

(B) The frequency $n$ of a vibrating wire is given by $n = \frac{1}{2 \ell} \sqrt{\frac{T}{m}}$.
Since the wires are identical,$\ell$ and $m$ are constant,so $n \propto \sqrt{T}$.
Let the initial frequency be $n_1$ and the initial tension be $T_1$. After increasing the tension by $2 \%$,the new tension $T_2 = T_1 + 0.02 T_1 = 1.02 T_1$.
The new frequency $n_2$ is given by $\frac{n_2}{n_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{1.02} = 1.01$.
Thus,$n_2 = 1.01 n_1$.
The beat frequency is given as $n_2 - n_1 = 5 \ Hz$.
Substituting $n_2$,we get $1.01 n_1 - n_1 = 5$.
$0.01 n_1 = 5$.
$n_1 = \frac{5}{0.01} = 500 \ Hz$.

(B) Let the frequency of tuning fork $C$ be $f_C$.
Given that the frequency of $A$ is $1.5 \%$ more than $f_C$:
$f_A = f_C + 0.015 f_C = 1.015 f_C$.
Given that the frequency of $B$ is $2.5 \%$ less than $f_C$:
$f_B = f_C - 0.025 f_C = 0.975 f_C$.
When $A$ and $B$ are sounded together, the beat frequency is $12 \text{ Hz}$:
$|f_A - f_B| = 12$.
Substituting the expressions:
$1.015 f_C - 0.975 f_C = 12$.
$0.040 f_C = 12$.
$f_C = \frac{12}{0.040} = 300 \text{ Hz}$.

(A) Let the frequency of tuning fork $A$ be $f_A$ and the frequency of tuning fork $B$ be $f_B = 480 \ Hz$.
The beat frequency is given by $|f_A - f_B| = 5 \ Hz$.
This implies $f_A = 480 \pm 5$,so $f_A$ is either $485 \ Hz$ or $475 \ Hz$.
When wax is added to tuning fork $A$,its frequency $f_A$ decreases.
After adding wax,the new beat frequency is $|f_A' - 480| = 2 \ Hz$,where $f_A' < f_A$.
If $f_A = 475 \ Hz$,adding wax would decrease the frequency further (e.g.,to $473 \ Hz$),making the beat frequency $|473 - 480| = 7 \ Hz$,which is not $2 \ Hz$.
If $f_A = 485 \ Hz$,adding wax decreases the frequency towards $480 \ Hz$,making the beat frequency $|482 - 480| = 2 \ Hz$.
Thus,the initial frequency of tuning fork $A$ was $485 \ Hz$.

(D) Beats are the periodic variations in the intensity of sound heard when two sound waves of slightly different frequencies and comparable amplitudes interfere with each other.
For the formation of distinct beats,the two sound sources must have nearly equal frequencies so that the beat frequency $(f_{beat} = |f_1 - f_2|)$ is low enough to be perceived by the human ear.
Additionally,they should have nearly equal amplitudes to ensure that the interference results in clearly audible maximums and minimums of intensity.

(A) The frequency of a vibrating wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,which implies $f \propto \sqrt{T}$.
Let $f_1$ be the frequency at tension $T_1 = 225 \ N$ and $f_2$ be the frequency at tension $T_2 = 256 \ N$.
Since $f \propto \sqrt{T}$,we have $\frac{f_1}{f_2} = \sqrt{\frac{225}{256}} = \frac{15}{16}$,so $f_2 = \frac{16}{15} f_1$.
Given that the beat frequency is $6 \ Hz$ in both cases,let $x$ be the frequency of the tuning fork.
For the first case: $f_1 = x - 6$ (assuming $x > f_1$).
For the second case: $f_2 = x + 6$ (since $f_2 > f_1$,the frequency must increase past the fork frequency).
Substituting $f_1$ and $f_2$: $x + 6 = \frac{16}{15} (x - 6)$.
$15(x + 6) = 16(x - 6) \implies 15x + 90 = 16x - 96$.
$x = 90 + 96 = 186 \ Hz$.

(C) Let the two waves have the same amplitude $A$ and intensities $I_0$. The intensity of a wave is proportional to the square of its amplitude,so $I_0 \propto A^2$.
When these two waves superpose,the maximum amplitude $A_{max}$ occurs during constructive interference,where $A_{max} = A + A = 2A$.
The maximum intensity $I_{max}$ is proportional to the square of the maximum amplitude: $I_{max} \propto (A_{max})^2 = (2A)^2 = 4A^2$.
Since $I_0 \propto A^2$,we have $I_{max} = 4I_0$.
Therefore,the maximum intensity is $4$ times the intensity of each constituent wave.

(C) The number of beats per second is given by the difference in frequencies,$|n_{1} - n_{2}|$.
Given the wave equations $y_{1} = a \sin(2000 \pi t)$ and $y_{2} = a \sin(2008 \pi t)$,we compare them with the standard form $y = a \sin(2 \pi n t)$.
For the first wave: $2 \pi n_{1} = 2000 \pi \implies n_{1} = 1000 \text{ Hz}$.
For the second wave: $2 \pi n_{2} = 2008 \pi \implies n_{2} = 1004 \text{ Hz}$.
The number of beats per second is $|n_{2} - n_{1}| = |1004 - 1000| = 4 \text{ beats per second}$.

(A) Beats are a phenomenon that occurs when two sound waves of slightly different frequencies,traveling in the same direction,superimpose on each other.
This superposition leads to the periodic variation in the intensity of the resultant sound,which is known as interference.
Therefore,the production of beats is a direct consequence of the interference of sound waves.

(C) The time period of one vibration is $T = \frac{1}{n} \ s$.
For $x$ vibrations,the total time taken is $t = x \times T = \frac{x}{n} \ s$.
The distance traveled by the wave in time $t$ is given by $d = V \times t$.
Substituting the value of $t$,we get $d = V \times \frac{x}{n} = \frac{xV}{n} \ m$.

(B) The distance between $N$ consecutive crests is given by $(N-1) \lambda$,where $\lambda$ is the wavelength.
Given that the distance between $5$ consecutive crests is $0.8 \ m$,we have:
$4 \lambda = 0.8 \ m$
$\lambda = \frac{0.8}{4} = 0.2 \ m$
Using the wave equation $v = n \lambda$,where $v$ is the velocity and $n$ is the frequency:
$n = \frac{v}{\lambda}$
Substituting the given values $v = 56 \ m/s$ and $\lambda = 0.2 \ m$:
$n = \frac{56}{0.2} = 280 \ Hz$
Therefore,the frequency of the tuning fork is $280 \ Hz$.

(B) Let the frequency of the $1^{\text{st}}$ tuning fork be $n_1$.
The frequencies are in an arithmetic progression with common difference $d = 5 \text{ Hz}$.
The frequency of the $41^{\text{st}}$ tuning fork is given by $n_{41} = n_1 + (41 - 1) \times d$.
$n_{41} = n_1 + 40 \times 5 = n_1 + 200$.
Given that the frequency of the last fork is double the first,$n_{41} = 2n_1$.
Substituting this into the equation: $2n_1 = n_1 + 200$.
Solving for $n_1$: $n_1 = 200 \text{ Hz}$.
Therefore,$n_{41} = 2 \times 200 = 400 \text{ Hz}$.

(D) The frequency of a sonometer wire is given by $f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$.
Since the tension $T$ and mass per unit length $\mu$ are constant,$f \propto \frac{1}{l}$.
Let $f_1$ and $f_2$ be the frequencies of the two tuning forks corresponding to lengths $l_1 = 0.97 \ m$ and $l_2 = 0.96 \ m$ respectively.
Thus,$f_1 = \frac{k}{0.97}$ and $f_2 = \frac{k}{0.96}$,where $k = \frac{1}{2} \sqrt{\frac{T}{\mu}}$.
Given the beat frequency is $f_2 - f_1 = 5 \ Hz$.
Substituting the values: $\frac{k}{0.96} - \frac{k}{0.97} = 5$.
$k \left( \frac{0.97 - 0.96}{0.96 \times 0.97} \right) = 5$.
$k \left( \frac{0.01}{0.9312} \right) = 5 \implies k = 5 \times 93.12 = 465.6$.
Now,$f_1 = \frac{465.6}{0.97} = 480 \ Hz$.
And $f_2 = \frac{465.6}{0.96} = 485 \ Hz$.

(B) Given: Beat frequency $= 4 \,Hz$, frequency of fork $B$ $(f_B)$ $= 384 \,Hz$.
The possible frequencies of fork $A$ $(f_A)$ are $f_B \pm 4$, which are $388 \,Hz$ or $380 \,Hz$.
When a prong of a tuning fork is filed, its mass decreases, which increases its frequency ($f_A$ increases).
Case $1$: If $f_A = 380 \,Hz$, filing increases $f_A$ towards $384 \,Hz$, so the beat frequency $(|f_A - f_B|)$ would decrease.
Case $2$: If $f_A = 388 \,Hz$, filing increases $f_A$ away from $384 \,Hz$ (e.g., to $389 \,Hz$), so the beat frequency $(|f_A - f_B|)$ increases.
Since the problem states that the beat frequency increases, the initial frequency of fork $A$ must be $388 \,Hz$.

(C) Let the unknown frequency be $n \,Hz$.
The beat frequency with $A$ $(n_A = 258 \,Hz)$ is $x = |n - 258|$.
The beat frequency with $B$ $(n_B = 262 \,Hz)$ is $2x = |n - 262|$.
If $n < 258$, then $x = 258 - n$.
Then $2x = |n - 262| = 262 - n$.
Substituting $x$: $2(258 - n) = 262 - n$.
$516 - 2n = 262 - n$.
$n = 516 - 262 = 254 \,Hz$.
Checking the condition: If $n = 254 \,Hz$, beats with $A$ is $|254 - 258| = 4 \,Hz$.
Beats with $B$ is $|254 - 262| = 8 \,Hz$.
Since $8 = 2 \times 4$, the condition is satisfied.

(D) The speed of sound $v = 330 \ m \ s^{-1}$.
The wavelengths are $\lambda_1 = 0.99 \ m$ and $\lambda_2 = 1.00 \ m$.
The frequencies are $f_1 = \frac{v}{\lambda_1} = \frac{330}{0.99} = \frac{33000}{99} = \frac{1000}{3} \ Hz$ and $f_2 = \frac{v}{\lambda_2} = \frac{330}{1.00} = 330 \ Hz$.
The beat frequency $f_b = |f_1 - f_2| = |\frac{1000}{3} - 330| = |\frac{1000 - 990}{3}| = \frac{10}{3} \ Hz$.
Beat frequency is defined as the number of beats per second. Thus,$f_b = \frac{\text{Number of beats}}{t}$.
Given that $10$ beats are produced in $t$ seconds,we have $\frac{10}{3} = \frac{10}{t}$.
Therefore,$t = 3 \ s$.

(D) The given equations of the two sound waves are $y_1 = 3 \sin(250 \pi t)$ and $y_2 = 2 \sin(260 \pi t)$.
Comparing these with the standard equation $y = A \sin(\omega t)$,we get the angular frequencies as $\omega_1 = 250 \pi \text{ rad/s}$ and $\omega_2 = 260 \pi \text{ rad/s}$.
The frequencies $f_1$ and $f_2$ are given by $f = \frac{\omega}{2 \pi}$.
Thus,$f_1 = \frac{250 \pi}{2 \pi} = 125 \text{ Hz}$ and $f_2 = \frac{260 \pi}{2 \pi} = 130 \text{ Hz}$.
The beat frequency $f_b$ is the difference between the two frequencies: $f_b = |f_2 - f_1| = |130 - 125| = 5 \text{ Hz}$.
The time interval between two successive maximum intensities is the time period of the beats,given by $T_b = \frac{1}{f_b}$.
Therefore,$T_b = \frac{1}{5} = 0.2 \text{ s}$.
Thus,the correct option is $D$.

(C) The beat frequency is given by $|n_A - n_B| = 4 \ Hz$.
When the tension in string $A$ is increased,its frequency $n_A$ increases because $n \propto \sqrt{T}$.
Since the number of beats increases from $4$ to $7$ upon increasing $n_A$,it implies that $n_A$ must be greater than $n_B$ (i.e.,$n_A - n_B = 4$).
If $n_A$ were less than $n_B$,increasing $n_A$ would decrease the beat frequency towards zero before increasing it again,but the problem implies a direct increase.
Therefore,$n_A = n_B + 4$.
Given $n_B = 480 \ Hz$,we have $n_A = 480 + 4 = 484 \ Hz$.

(D) The beat frequency is defined as the difference between the two frequencies,given by $f_{beat} = v_1 - v_2$.
The time interval between two successive maxima or two successive minima is known as the time period of the beats $(T_{beat})$.
The time period is the reciprocal of the beat frequency:
$T_{beat} = \frac{1}{f_{beat}} = \frac{1}{v_1 - v_2}$.
Therefore,the duration of time between two successive minima is $\frac{1}{v_1 - v_2}$.

(D) Beats are a phenomenon that occurs due to the superposition of two sound waves.
When two sound waves of equal amplitude and slightly different frequencies travel in the same direction,they interfere with each other.
This interference results in a periodic variation in the intensity of the resultant sound,which is perceived as beats.
Therefore,the formation of beats is a direct consequence of the interference of sound waves.

(D) Given, frequencies of tuning forks $X$ and $Y$ are:
$n_X = 280 \,Hz$
$n_Y = 284 \,Hz$
Let the frequency of tuning fork $Z$ be $n_Z$ and the beat frequency produced by $X$ and $Z$ be $b$.
From the first condition, $b = |n_Z - 280|$.
From the second condition, the beat frequency produced by $Y$ and $Z$ is $3b$, so $3b = |n_Z - 284|$.
Case $1$: If $n_Z > 284$, then $n_Z - 280 = b$ and $n_Z - 284 = 3b$. Substituting $b$, we get $n_Z - 284 = 3(n_Z - 280) \Rightarrow n_Z - 284 = 3n_Z - 840 \Rightarrow 2n_Z = 556 \Rightarrow n_Z = 278 \,Hz$. This contradicts $n_Z > 284$.
Case $2$: If $n_Z < 280$, then $280 - n_Z = b$ and $284 - n_Z = 3b$. Substituting $b$, we get $284 - n_Z = 3(280 - n_Z) \Rightarrow 284 - n_Z = 840 - 3n_Z \Rightarrow 2n_Z = 556 \Rightarrow n_Z = 278 \,Hz$. This is consistent with $n_Z < 280$.
Thus, the frequency of $Z$ is $278 \,Hz$.

(A) Let the frequency of the first tuning fork be $f_1 = f$.
The number of tuning forks is $n = 56$.
The beat frequency is $d = 4 \text{ Hz}$.
The frequency of the $n^{\text{th}}$ fork is given by $f_n = f_1 + (n - 1)d$.
So, $f_{56} = f + (56 - 1) \times 4 = f + 55 \times 4 = f + 220$.
Given that the frequency of the last fork is twice that of the first, we have $f_{56} = 2f$.
Equating the two expressions: $2f = f + 220$, which gives $f = 220 \text{ Hz}$.
The frequency of the $19^{\text{th}}$ fork is $f_{19} = f_1 + (19 - 1)d$.
$f_{19} = 220 + 18 \times 4 = 220 + 72 = 292 \text{ Hz}$.

(A) The beat frequency is given by the difference between the two frequencies: $f_{beat} = |f_2 - f_1| = |323 \ Hz - 320 \ Hz| = 3 \ Hz$.
This means there are $3$ beats per second.
The time interval between two successive beats (one maximum and one adjacent minimum) is half of the time period of one beat cycle.
The time period of the beat is $T = \frac{1}{f_{beat}} = \frac{1}{3} \ s$.
The time interval between a maximum sound and its adjacent minimum sound is $\Delta t = \frac{T}{2} = \frac{1}{2} \times \frac{1}{3} \ s = \frac{1}{6} \ s$.

(C) Given frequencies are $f_1 = n-1$,$f_2 = n$,and $f_3 = n+1$.
Beats are produced due to the difference in frequencies of the sound sources.
The number of beats produced per second is determined by the difference between the maximum and minimum frequencies present in the system.
$\text{Beats produced} = f_{\text{max}} - f_{\text{min}} = (n+1) - (n-1) = 2 \text{ beats/sec}$.
Since all three sources are vibrating together,the interference pattern results in a beat frequency of $2 \text{ Hz}$.
Therefore,the number of beats produced is $2$ and the number of beats heard is also $2$.
Thus,the correct option is $C$.

(D) Given equations are $y_1 = 5 \sin(400 \pi t)$ and $y_2 = 8 \sin(408 \pi t)$.
Comparing these with the standard equation $y = A \sin(\omega t)$,we get angular frequencies $\omega_1 = 400 \pi \text{ rad/s}$ and $\omega_2 = 408 \pi \text{ rad/s}$.
The frequencies $f_1$ and $f_2$ are given by $f = \frac{\omega}{2 \pi}$.
$f_1 = \frac{400 \pi}{2 \pi} = 200 \text{ Hz}$ and $f_2 = \frac{408 \pi}{2 \pi} = 204 \text{ Hz}$.
The beat frequency is $|f_2 - f_1| = |204 - 200| = 4 \text{ beats per second}$.
To find the number of beats per minute,we multiply by $60$:
$\text{Beats per minute} = 4 \times 60 = 240 \text{ beats/min}$.

(A) Given: Frequency of tuning fork $A$ $(f_A)$ = $250 \,Hz$. Frequency of tuning fork $B$ $(f_B)$ = $x \,Hz$.
Initially, the beat frequency is $5 \,Hz$, so $|f_A - f_B| = 5$.
This implies $250 - x = 5$ or $x - 250 = 5$, giving $x = 245 \,Hz$ or $x = 255 \,Hz$.
When tuning fork $B$ is waxed, its frequency decreases $(f_B' < f_B)$.
After waxing, the new beat frequency is $3 \,Hz$.
If $x = 255 \,Hz$, then $f_B$ decreases towards $250 \,Hz$, so the beat frequency would increase from $5$ to a higher value or pass through $0$ and then increase. This does not match the observation of $3 \,Hz$.
If $x = 245 \,Hz$, then $f_B$ decreases further away from $250 \,Hz$ (e.g., to $244 \,Hz$), which would increase the beat frequency to $6 \,Hz$. However, if we consider the frequency of $B$ was $245 \,Hz$ and it was slightly loaded, the beat frequency decreases from $5$ to $3$ only if the frequency of $B$ was originally $255 \,Hz$ and it decreased to $253 \,Hz$ ($255 - 253 = 2$ - incorrect) or if $x = 245 \,Hz$ and the beat frequency changed. Let's re-evaluate: If $x = 255 \,Hz$, loading $B$ makes $f_B$ approach $250 \,Hz$, decreasing the beat frequency. Thus, $x = 255 \,Hz$ is the correct value.

(A) Let the frequency of string $A$ be $f_A = f_0$ and the frequency of string $B$ be $f_B$.
Initially,the beat frequency is $\Delta f_1 = |f_0 - f_B| > 0$.
When the tension in string $A$ is increased,its frequency $f_A$ increases to $f_A'$.
The new beat frequency is $\Delta f_2 = |f_A' - f_B| < \Delta f_1$.
Since the beat frequency decreased after increasing the frequency of $A$,it implies that $f_A$ was approaching $f_B$.
Therefore,$f_B$ must be greater than $f_A$.
Thus,$\Delta f_1 = f_B - f_0$,which gives $f_B = f_0 + \Delta f_1$.

(A) The fundamental frequency of a stretched wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Since the wires are identical,$L$ and $\mu$ are constant,so $f \propto \sqrt{T}$.
Initially,for both wires,the frequency is $f_0 \propto \sqrt{T}$. Thus,$f_0^2 \propto T$ ... $(i)$
When the tension of one wire is increased by $\Delta T$,its new frequency becomes $f' = f_0 + N$.
Therefore,$(f_0 + N)^2 \propto (T + \Delta T)$ ... (ii)
Dividing equation (ii) by equation $(i)$,we get:
$\frac{(f_0 + N)^2}{f_0^2} = \frac{T + \Delta T}{T}$
$\frac{(f_0 + N)^2}{f_0^2} = 1 + \frac{\Delta T}{T}$
Rearranging for $\frac{\Delta T}{T}$:
$\frac{\Delta T}{T} = \frac{(f_0 + N)^2}{f_0^2} - 1 = \left(\frac{f_0 + N}{f_0}\right)^2 - 1$.

(A) Let the velocity of sound be $v$ and the frequency of the third note be $f_0$. The frequencies of the two given notes are $f_1 = \frac{v}{\lambda_1} = \frac{v}{40/195} = \frac{195v}{40}$ and $f_2 = \frac{v}{\lambda_2} = \frac{v}{40/193} = \frac{193v}{40}$.
Since each note produces $9$ beats per second with the third note,we have:
$|f_1 - f_0| = 9$ and $|f_2 - f_0| = 9$.
This implies $f_1 - f_0 = 9$ and $f_0 - f_2 = 9$ (assuming $f_1 > f_0 > f_2$).
Adding these two equations:
$(f_1 - f_0) + (f_0 - f_2) = 9 + 9$
$f_1 - f_2 = 18$
$\frac{195v}{40} - \frac{193v}{40} = 18$
$\frac{2v}{40} = 18$
$\frac{v}{20} = 18$
$v = 360 \,m/s$.