Question

1
If the length of a closed organ pipe is 1 m and velocity of
sound is 330 m/s, then the frequency of 1st overtone is :
(a) 4 (330/4) Hz (b)3 (330/4) Hz
(c)2 (330/4) Hz
(d) none of these​

Answer 1

Frequency of the 1st overtone in a closed organ pipe is 3 (330/4) Hz.

Option (b)

Explanation:

The equation for the fundamental frequency in a closed organ pipe is

$\eta_{=\frac{V}{4 L}}$  .... (Eq 1)

Where, V is the velocity of sound

L is the Length of the organ pipe

It is given that the length of the pipe (L) is 1 m and Velocity of the sound (V) is 330 m/s.

Substituting this in equation 1, we get

$\eta=\frac{330}{4}$ ....(eq 2)

For the first overtone,

The frequency is given by $F=3 \eta$ .... (Eq 3).

From the equation 2, Substitute the value of $\eta$ in equation 3.

$F=3 \frac{330}{4} \mathrm{Hertz}$

Thus for the closed organ pipe the frequency of the first overtone is $3\frac{330}{4}$ Hz.

To learn more

Find the value of a and prepare a frequency distribution table for drawing a frequency polygon. draw a frequency polygon.

class :-20-30,30-40,40-50,50-60,

Frequency :-a, 2a, 3a, a, total 70​

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