### Waves - Result Question 37

#### 37. In a guitar, two strings $A$ and $B$ made of same material are slightly out of tune and produce beats of frequency $6 Hz$. When tension in B is slightly decreased, the beat frequency increases to $7 Hz$. If the frequency of A is $530 Hz$, the original frequency of B will be

(a) $524 Hz$

(b) $536 Hz$

(c) $537 Hz$

(d) $523 Hz$

[2020]

## Show Answer

#### Answer:

Correct Answer: 37. (a)

Solution:

- (a) Frequency of string, $f=\frac{1}{2 l} \sqrt{\frac{T}{m}}$

Frequency $\propto \sqrt{\text{ Tension }}$

Difference of $f_A$ and $f_B$ is $6 Hz$.

If tension decreases, $f_B$ decreases and becomes

$f^{\prime} _{B}$.

Now, difference of $f_A$ and $f^{\prime} _{B}=7 Hz$ (increases)

So, $f_A>f_B$

$f_A-f_B=6 Hz$

$\Rightarrow f_A=530 Hz \Rightarrow f_B=524 Hz$ (original)

38 .

$ \text{ (d) } \begin{align*} l_1 & =9.75 cm \\ l_2 & =31.25 cm \\ l_3 & =52.75 cm \\ e & =\text{ end correction } \\ \frac{\lambda}{4}+e & =9.75 cm \tag{i}\\ \frac{3 \lambda}{4} & +e=31.25 cm \tag{ii}\\ \frac{3 \lambda}{4} & -\frac{\lambda}{4}=31.25-9.75 \\ \frac{\lambda}{2} & =21.5 \\ \lambda & =43 cm^{2} \\ v & =\lambda \times f \\ v & =34400 \times 10^{-2} \\ \therefore \quad v & =344 ms^{-1} \end{align*} . $