Search any question & find its solution
Question:
Answered & Verified by Expert
The fundamental frequency of open pipe is 'n'. If it is closed from one end then frequency of the $2^{\text {nd }}$ harmonic of closed pipe is higher by $200 \mathrm{~Hz}$ than ${ }^{\prime} \mathrm{n}^{\prime}$. The value of 'n' is
Options:
Solution:
1344 Upvotes
Verified Answer
The correct answer is:
$400 \mathrm{~Hz}$
$\mathrm{n}=\frac{\mathrm{v}}{2 \ell}$
$2^{\mathrm{nd}}$ harmonic $=1^{\mathrm{st}}$ overtone $=(2 \mathrm{p}+1) \mathrm{n}_{0}=3 \mathrm{n}_{0}$
$\frac{3 \mathrm{v}}{4 \ell}-\frac{\mathrm{v}}{2 \ell}=200$
$\frac{v}{2 \ell}\left(\frac{3}{2}-1\right)=200$
$\frac{v}{2 \ell}=400$
$2^{\mathrm{nd}}$ harmonic $=1^{\mathrm{st}}$ overtone $=(2 \mathrm{p}+1) \mathrm{n}_{0}=3 \mathrm{n}_{0}$
$\frac{3 \mathrm{v}}{4 \ell}-\frac{\mathrm{v}}{2 \ell}=200$
$\frac{v}{2 \ell}\left(\frac{3}{2}-1\right)=200$
$\frac{v}{2 \ell}=400$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.