Question

2 organ pipes 80cm and 81 cm long are found to give beat frequency of 2.6 Hz when each is sounding it's fundamental note ,neglecting end correction . calculate
1: the velocity of sound in air
2: frequency of the 2 notes

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

The velocity of sound in air is 337m/s, and the frequency of the 2 notes are 210.6Hz and 208 Hz respectively.

Explanation:

The given organ pipes are of open type.

And for an open organ pipe, the fundamental frequency is given as,

$f=\frac{v}{2l}$        (1)

Where,

f=frequency of an organ pipe

v=velocity of sound in air

l=length of an organ pipe

From the question we have,

l₁=80cm=0.8m

l₂=81cm=0.81m

Beat frequency=2.6Hz

And the beat frequency is given as,

$f_1-f_2=2.6Hz$       (2)

f₁=fundamental frequency of first organ pipe

f₂=fundamental frequency of second organ pipe

By solving equation (2) we get;

$\frac{v}{2l_1} -\frac{v}{2l_2} =2.6$      (3)

By substituting the value of l₁ and l₂ in equation (3) we get;

$\frac{v}{2\times 0.8} -\frac{v}{2\times 0.81} =2.6$

$\frac{v}{2} (\frac{0.81-0.8}{0.8\times 0.81})=2.6$

$v=\frac{2.6\times 2\times 0.8\times 0.81}{0.01}=337m/s$        (4)

The frequency of the first note is,

$f_1=\frac{337}{2\times 0.8}=210.6Hz$        (5)

The frequency of the second note is,

$f_2=\frac{337}{2\times 0.81}=208Hz$       (6)

Hence, the velocity of sound in air is 337m/s, and the frequency of the 2 notes are 210.6Hz and 208 Hz respectively.