Question

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Answer 1

Answer:

128Hz

Explanation:

Since this is a closed pipe, the frequencies are given by

$\boxed{v=\frac{(2n+1)V}{4L}}$

Therefore two consecutive frequencies differ by a factor k, given by:

$k=\frac{(2n+1)V}{4L}-\frac{(2(n-1)+1)V}{4L}\\\\k=\frac{2V}{4L}\\\\\boxed{k=\frac{V}{2L}}$

Difference in the given overtones are :-

$=>896-640=256Hz\\\\=>1152-896=256Hz$

Since the difference in these overtones is same, this implies that these are consecutive overtones

∴ $k=\boxed{\frac{V}{2L}=256Hz}$

Since Fundamental frequency of a closed pipe is $\boxed{f=\frac{V}{4L}}$

$=>f=k/2\\\\=>f=\frac{256}{2}\\\\=>\boxed{\boxed{f=128 Hz}}$

Hope this answer helped you :)