Q. 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
A) $4 \times \frac{330}{4}Hz$
B) $3 \times \frac{330}{4}Hz$
C) $2 \times \frac{330}{4}Hz$
D) $2 \times \frac{4}{330}Hz$
Explanation
(b) For closed pipe $n_{1} = \frac{v}{4l} = \frac{330}{4}Hz$
Second note $= 3n_{1} = \frac{3 \times 330}{4}Hz$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.