Wave Mechanics Question 650
Question: The length of the wire between two ends of asonometer is 100 cm. What should be the positions of two bridges below the wire so that the three segments of the wire have their fundamental frequencies in the ratio of 1 : 3 : 5?
Options:
A) $ \frac{1500}{23}cm,\frac{2000}{23}cm $
B) $ \frac{1500}{23}cm,\frac{500}{23}cm $
C) $ \frac{1500}{23}cm,\frac{300}{23}cm $
D) $ \frac{300}{23}cm,\frac{1500}{23}cm $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] From formula, $ f=\frac{1}{x}\sqrt{\frac{T}{m}}\Rightarrow \frac{1}{f}\propto l $
$ \
Therefore l _{1}:l _{2}:l _{3}=\frac{1}{f _{1}}:\frac{1}{f _{2}}:\frac{1}{f _{3}}=f _{2}f _{3}:f _{1}f _{3}:f _{1}f _{2} $
$ [ Given:f _{1}:f _{2}:f _{3}=1:3:5 ] $ =15: 5: 3
Therefore the positions of two bridges below the wire are $ \frac{15\times 100}{15+5+3}cm,and,\frac{15\times 100+5\times 100}{15+5+3}cm $ i.e., $ \frac{1500}{23}cm,\frac{2000}{23}cm $
BETA
