Wave Mechanics Question 1114
Question: If the length of a closed organ pipe is 1m and velocity of sound is 330 m/s, then the frequency for the second note is
[AFMC 2001]
Options:
A) $ 4\times \frac{330}{4}Hz $
B) $ 3\times \frac{330}{4}Hz $
C) $ 2\times \frac{330}{4}Hz $
D) $ 2\times \frac{4}{330}Hz $
Show Answer
Answer:
Correct Answer: B
Solution:
For closed pipe $ n _{1}=\frac{v}{4l}=\frac{330}{4} $
$ Hz $ Second note = $ 3n _{1}=\frac{3\times 300}{4} $
$ Hz $ .
BETA
