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

• A. $4 \times \frac{330}{4}Hz$
• B. $3 \times \frac{330}{4}Hz$
• C. $2 \times \frac{330}{4}Hz$
• D. $2 \times \frac{4}{330}Hz$

Answer 1

Answer: (B)

$3 \times \frac{330}{4}Hz$

Answer 2

For closed pipe $n_{1} = \frac{v}{4l} = \frac{330}{4}Hz$

Second note = $3n_{1} = \frac{3 \times 300}{4}Hz$