Beats and Tuning fork Questions in English

(D) Let the frequency of the third note be $n$ and the velocity of sound be $v$.
The frequencies of the two notes are $n_1 = \frac{v}{\lambda_1} = \frac{v}{36/195} = \frac{195v}{36}$ and $n_2 = \frac{v}{\lambda_2} = \frac{v}{36/193} = \frac{193v}{36}$.
Since each note produces $10$ beats per second with the third note, we have:
$\frac{195v}{36} - n = 10$ ---$(i)$
$n - \frac{193v}{36} = 10$ ---(ii)
Adding equations $(i)$ and (ii):
$\frac{195v}{36} - \frac{193v}{36} = 10 + 10$
$\frac{2v}{36} = 20$
$\frac{v}{18} = 20$
$v = 360 \,m/s$.

(D) For a tube closed at one end, the fundamental frequency is given by $f_c = \frac{v}{4L}$.
Given that the beat frequency is $f_{c1} - f_{c2} = 2$, where $f_{c1} = \frac{v}{4L_1}$ and $f_{c2} = \frac{v}{4L_2}$.
Thus, $\frac{v}{4L_1} - \frac{v}{4L_2} = 2$.
For a tube open at both ends, the fundamental frequency is given by $f_o = \frac{v}{2L}$.
The new beat frequency is $f_{o1} - f_{o2} = \frac{v}{2L_1} - \frac{v}{2L_2}$.
We can rewrite this as $2 \times (\frac{v}{4L_1} - \frac{v}{4L_2})$.
Substituting the known value, the new beat frequency is $2 \times 2 = 4$ beats per second.

(A) Given: $\lambda_1 = 5 \,m$, $\lambda_2 = 6 \,m$.
The frequency of a wave is given by $n = \frac{v}{\lambda}$, where $v$ is the velocity of sound.
Frequency of the first wave: $n_1 = \frac{v}{5}$.
Frequency of the second wave: $n_2 = \frac{v}{6}$.
The beat frequency is the difference between the two frequencies: $n_1 - n_2 = \frac{\text{Number of beats}}{\text{Time}} = \frac{30}{3} = 10 \,Hz$.
Substituting the expressions for $n_1$ and $n_2$: $\frac{v}{5} - \frac{v}{6} = 10$.
Taking $v$ as a common factor: $v \left( \frac{6 - 5}{30} \right) = 10$.
$\frac{v}{30} = 10$.
Therefore, $v = 300 \,m/s$.

(B) The frequency of string $A$ is $f_A = 320 \ Hz$. Let the frequency of string $B$ be $f_B$. The beat frequency is given by $n = |f_A - f_B| = 8 \ Hz$.
This implies $f_B = 320 \pm 8$, so $f_B$ could be $312 \ Hz$ or $328 \ Hz$.
When the tension in string $A$ is reduced, its frequency $f_A$ decreases because $f \propto \sqrt{T}$.
Let the new frequency be $f_A'$. Since $f_A' < 320 \ Hz$, the new beat frequency is $n' = |f_A' - f_B| = 4 \ Hz$.
If $f_B = 312 \ Hz$, then $f_A' - 312 = 4 \implies f_A' = 316 \ Hz$ (which is less than $320 \ Hz$, consistent).
If $f_B = 328 \ Hz$, then $328 - f_A' = 4 \implies f_A' = 324 \ Hz$ (which is greater than $320 \ Hz$, inconsistent).
Therefore, the frequency of $B$ must be $312 \ Hz$.

(D) Given that, the fundamental frequency of the first wire is $f_1 = 500 \,Hz$.
Let the frequency of the second wire be $f_2$.
The beat frequency is $f_b = 5 \,Hz$.
When the tension in the first wire is decreased, its frequency decreases. Since it produces $5$ beats per second with the second wire, the frequency of the first wire must have been higher than the second wire initially, or the second wire is higher. Given the tension in the first wire is decreased, the new frequency $f_1' = f_1 - 5 = 495 \,Hz$.
However, the standard interpretation for this problem is that the second wire has a fixed frequency $f_2 = 505 \,Hz$ (or $495 \,Hz$).
Using the relation $f \propto \sqrt{T}$, we have $\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1}}$.
Taking $f_2 = 505 \,Hz$ and $f_1 = 500 \,Hz$, we get $\frac{T_2}{T_1} = (\frac{505}{500})^2 = (1.01)^2 = 1.0201 \approx 1.02$.

(B) Let the initial frequency of both wires be $n = 600 \text{ Hz}$.
The frequency of a stretched wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$.
When the tension in one wire is increased to $T'$, its new frequency becomes $n' = \frac{1}{2l} \sqrt{\frac{T'}{m}}$.
Given that the beat frequency is $6 \text{ Hz}$, we have $n' - n = 6$.
So, $n' = 600 + 6 = 606 \text{ Hz}$.
Taking the ratio of the two frequencies:
$\frac{n'}{n} = \frac{\frac{1}{2l} \sqrt{\frac{T'}{m}}}{\frac{1}{2l} \sqrt{\frac{T}{m}}} = \sqrt{\frac{T'}{T}}$
$\frac{606}{600} = \sqrt{\frac{T'}{T}}$
$1.01 = \sqrt{\frac{T'}{T}}$
Squaring both sides:
$\frac{T'}{T} = (1.01)^2 = 1.0201 \approx 1.02$.
The fractional increase in tension is $\frac{\Delta T}{T} = \frac{T' - T}{T} = \frac{T'}{T} - 1$.
$\frac{\Delta T}{T} = 1.02 - 1 = 0.02$.

(D) The phenomenon where the intensity of sound varies periodically in time is known as $beats$.
$Beats$ occur when two sound waves of slightly different frequencies, $f_1$ and $f_2$, interfere with each other.
The resulting intensity varies with a frequency equal to the difference of the two frequencies, $|f_1 - f_2|$.
Additionally, if the source intensity itself is modulated periodically, the observer will perceive a periodic change in intensity.
Comparing this with the given options, option $D$ describes the physical condition for $beats$, which is a standard cause for periodic intensity variation.

(A) The beat frequency is the number of beats per second. Initially, $8$ beats in $2$ s means the beat frequency is $f_{beat} = 8/2 = 4$ Hz.
This implies $|f_A - f_B| = 4$ Hz.
Given $f_B = 380$ Hz, so $|f_A - 380| = 4$, which means $f_A = 384$ Hz or $f_A = 376$ Hz.
When tuning fork $A$ is loaded with wax, its frequency $f_A$ decreases.
After loading, the new beat frequency is $4$ beats in $2$ s,which is $f'_{beat} = 4/2 = 2$ Hz.
If $f_A$ was $376$ Hz, loading it would decrease it further away from $380$ Hz, increasing the beat frequency.
If $f_A$ was $384$ Hz, loading it would decrease it towards $380$ Hz, reducing the beat frequency to $2$ Hz.
Therefore, the original frequency of tuning fork $A$ must be $384$ Hz.

(B) The beat frequency is given by $|n - f_s| = x$, where $f_s$ is the frequency of the sonometer string. This implies $f_s = n + x$ or $f_s = n - x$.
When the tension $T$ of the string is increased, the frequency of the string $f_s$ increases because $f_s \propto \sqrt{T}$.
If $f_s = n + x$, increasing $f_s$ will make it move further away from $n$, thus increasing the beat frequency $(f_s - n)$.
If $f_s = n - x$, increasing $f_s$ will make it move closer to $n$, thus decreasing the beat frequency $(n - f_s)$.
Since the problem states that the beat frequency decreases, the initial frequency of the string must have been $n - x$.

(A) The general equation of a wave is given by $Y = A \sin(\omega t)$.
Comparing the given equations with the standard form:
For $Y_1 = 0.25 \sin 316t$, the angular frequency is $\omega_1 = 316 \text{ rad/s}$.
For $Y_2 = 0.25 \sin 310t$, the angular frequency is $\omega_2 = 310 \text{ rad/s}$.
The angular frequencies are related to linear frequencies by $\omega = 2\pi f$, so $f = \omega / (2\pi)$.
Therefore, $f_1 = 316 / (2\pi)$ and $f_2 = 310 / (2\pi)$.
The beat frequency is the difference between the two frequencies: $f_b = |f_1 - f_2|$.
$f_b = |(316 / 2\pi) - (310 / 2\pi)| = 6 / (2\pi) = 3 / \pi$.
Thus, the number of beats produced per second is $3/\pi$.

(C) Beats are a phenomenon in acoustics that occur due to the superposition of two sound waves of slightly different frequencies traveling in the same direction.
When these waves superimpose, they create a periodic variation in the intensity of sound at a point, which is heard as a waxing and waning of loudness.
For the phenomenon of beats to be clearly audible, the frequencies of the two waves must be very close to each other (i.e.,the beat frequency $f_b = |f_1 - f_2|$ should be small, typically $\le 10 \text{ Hz}$).
While the amplitudes do not strictly need to be the same, the effect is most pronounced when the amplitudes are equal or nearly equal.

(D) The given wave equations are $y_1 = 4 \sin(708 \pi t)$ and $y_2 = 3 \sin(700 \pi t)$.
Comparing these with the standard equation $y = A \sin(2 \pi f t)$, we get:
For $y_1$: $2 \pi f_1 = 708 \pi \implies f_1 = 354 \text{ Hz}$ and amplitude $A_1 = 4$.
For $y_2$: $2 \pi f_2 = 700 \pi \implies f_2 = 350 \text{ Hz}$ and amplitude $A_2 = 3$.
Beat frequency $f_b = |f_1 - f_2| = |354 - 350| = 4 \text{ beats per second}$.
The intensity $I$ is proportional to the square of the amplitude $(I \propto A^2)$.
Maximum intensity (waxing) $I_{max} \propto (A_1 + A_2)^2 = (4 + 3)^2 = 7^2 = 49$.
Minimum intensity (wanning) $I_{min} \propto (A_1 - A_2)^2 = (4 - 3)^2 = 1^2 = 1$.
Therefore, the ratio of intensity between waxing and wanning is $I_{max} : I_{min} = 49:1$.
The number of beats per second is $4$ and the intensity ratio is $49:1$.

(C) Let the frequencies of the $14$ tuning forks be in an arithmetic progression: $f_1, f_2, f_3, ..., f_{14}$.
Given that each fork produces '$x$' beats per second with the preceding one, the common difference is $d = x$.
Thus, $f_n = f_1 + (n-1)x$.
The last fork is an octave of the first, meaning $f_{14} = 2f_1$.
Substituting $n=14$: $f_1 + 13x = 2f_1$, which implies $f_1 = 13x$.
The frequency of the seventh fork is $f_7 = f_1 + 6x = 114 \text{ Hz}$.
Substituting $f_1 = 13x$ into the equation: $13x + 6x = 114$.
$19x = 114$.
$x = 114 / 19 = 6$.
Therefore, the value of $x$ is $6$.

(C) For a pipe open at both ends, the fundamental frequency is given by $f = \frac{v}{2l}$, where $v$ is the velocity of sound and $l$ is the length of the pipe.
For the first pipe of length $L$, the frequency is $f_1 = \frac{v}{2L}$.
For the second pipe of length $(L + L_1)$, the frequency is $f_2 = \frac{v}{2(L + L_1)}$.
The beat frequency is the difference between the two frequencies: $f_b = |f_1 - f_2|$.
$f_b = \left| \frac{v}{2L} - \frac{v}{2(L + L_1)} \right| = \frac{v}{2} \left| \frac{(L + L_1) - L}{L(L + L_1)} \right|$.
$f_b = \frac{v}{2} \left( \frac{L_1}{L(L + L_1)} \right) = \frac{vL_1}{2L(L + L_1)}$.

(C) tuning fork resonates with frequencies that are integer multiples of its fundamental frequency '$n$'.
These are known as harmonics or overtones.
The given fundamental frequency is $n = 288$ Hz.
The resonant frequencies are given by $f = k \times n$, where $k = 1, 2, 3, \dots$
For $k = 1$, $f = 1 \times 288 = 288$ Hz.
For $k = 2$, $f = 2 \times 288 = 576$ Hz.
For $k = 3$, $f = 3 \times 288 = 864$ Hz.
Comparing these with the given options:
Option $A$ ($288$ Hz) is the fundamental frequency $(k=1)$.
Option $B$ ($576$ Hz) is the second harmonic $(k=2)$.
Option $D$ ($864$ Hz) is the third harmonic $(k=3)$.
Option $C$ ($844$ Hz) is not an integer multiple of $288$ Hz $(844 / 288 \approx 2.93)$.
Therefore, the tuning fork will not resonate with $844$ Hz.

(A) The fundamental frequency of wire $A$ is $f_A = 256 \text{ Hz}$.
The third harmonic of wire $A$ is $3 \times f_A = 3 \times 256 = 768 \text{ Hz}$.
The fundamental frequency of wire $B$ is $f_B = 382 \text{ Hz}$.
The second harmonic of wire $B$ is $2 \times f_B = 2 \times 382 = 764 \text{ Hz}$.
The beat frequency is the absolute difference between the two frequencies: $|768 - 764| = 4 \text{ Hz}$.
This means $4$ beats are heard per second.
Therefore, in $2$ seconds, the number of beats heard will be $4 \times 2 = 8$.

(B) Step $1$: From the first condition, the beat frequency is $|\nu - 200| = 6$. This implies $\nu = 200 \pm 6$, so $\nu = 206 \text{ Hz}$ or $\nu = 194 \text{ Hz}$.
Step $2$: From the second condition, the beat frequency for the second harmonic $2\nu$ is $|2\nu - 420| = 8$. This implies $2\nu = 420 \pm 8$, so $2\nu = 428 \text{ Hz}$ or $2\nu = 412 \text{ Hz}$.
Step $3$: Solving for $\nu$ in the second condition gives $\nu = 214 \text{ Hz}$ or $\nu = 206 \text{ Hz}$.
Step $4$: Comparing the results from both conditions, the common value is $\nu = 206 \text{ Hz}$.