Search any question & find its solution
Question:
Answered & Verified by Expert
A string is stretched between fixed points separated by $75 \mathrm{~cm}$. It is observed to have resonant frequencies of $420 \mathrm{~Hz}$ and $315 \mathrm{~Hz}$. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is
Options:
Solution:
1611 Upvotes
Verified Answer
The correct answer is:
$105 \mathrm{~Hz}$
$105 \mathrm{~Hz}$
$\frac{\mathrm{n}}{2 \ell}(\mathrm{v})=315, \frac{(\mathrm{n}+1)}{2 \ell} \mathrm{v}=420$
Solving $\frac{\mathrm{v}}{2 \ell}=105$
Solving $\frac{\mathrm{v}}{2 \ell}=105$
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.