Waves - Result Question 58
60. Each of the two strings of length $51.6 cm$ and $49.1 cm$ are tensioned separately by $20 N$ force. Mass per unit length of both the strings is same and equal to $1 g / m$. When both the strings vibrate simultaneously the number of beats is
[2009]
(a) 7
(b) 8
(c) 3
(d) 5
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Answer:
Correct Answer: 60. (a)
Solution:
- (a) The frequency of vibration of a string is given by, $f=\frac{1}{21} \sqrt{\frac{T}{m}}$ where $m$ is mass per unit length.
$f_1=\frac{1}{2 l_1} \sqrt{\frac{T}{m}}, f_2=\frac{1}{2 l_2} \sqrt{\frac{T}{m}}$,
$f_2-f_1=\frac{1}{2} \sqrt{\frac{T}{m}} \frac{(l_1-l_2)}{l_1 l_2}$
$\sqrt{\frac{T}{m}}=\sqrt{\frac{20}{10^{-3}}}=\sqrt{2} \times 10^{2}=1.414 \times 100$
$=141.4$
$\frac{1_1-1_2}{1_1 l_2}=\frac{(51.6-49.1) \times 10^{2}}{51.6 \times 49.1}$
$=\frac{2.5 \times 10^{2}}{50 \times 50}=\frac{1}{10}$
$\therefore \quad f_2-f_1=\frac{1}{2} \times 141.4 \times \frac{1}{10}=7$ beats
