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.