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]
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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*} . $
