Question

A pipe closed at one end and open at the other end resonates
with sound waves of frequencies 135 Hz and also 165 Hz
but not with any wave of frequency intermediate between
these two. The frequency of the fundamental note is :
(a) 30 Hz
(b) 15 Hz
(c) 60 Hz
(d) 7.5 Hz
tal​

Answer 1

Answer:

b)15Hz

Explanation:

least common multiple of 135,165

find HCF 135=3×3×3×5

165=3×5×11

HCF 3×5=15

Answer 2

Fundamental frequency is 15.

We know , frequency of sound wave in pipe closed at one end and open at the other end resonates, $f=\dfrac{(2n+1)v}{4L}$.

Here, v = velocity of speed.

         L = length of tube.

Now, it is given that their is no any wave between 135 Hz and 165 Hz.

Therefore, these are consecutive resonating frequencies.

So,

$135=\dfrac{(2n+1)v}{4L}\\165=\dfrac{(2(n+1)+1)v}{4L}=\dfrac{(2n+3)v}{4L}$

Subtracting them we get ,

$\\\\165-135=\dfrac{(2n+3)v}{4L}-\dfrac{(2n+1)v}{4L}\\30=\dfrac{2v}{4L}\\\dfrac{v}{4L}=15$

Therefore, fundamental frequency is 15.

Hence, this is the required solution.

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