Wave Mechanics Question 658
Question: A string is stretched between fixed points separated by 75.0 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is
Options:
A) 105 Hz
B) 1.05 Hz
C) 1050 Hz
D) 10.5 Hz
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Given $ \frac{nv}{2\ell }=315and,(n+1)\frac{v}{2\ell }=420 $
$ \Rightarrow \frac{n+1}{n}=\frac{420}{315}\Rightarrow n=3 $ Hence $ 3\times \frac{v}{2\ell }=315\Rightarrow \frac{v}{2\ell }=105Hz $ The lowest resonant frequency is when n = 1
Therefore lowest resonant frequency = 105Hz.
BETA
