Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
A string of length $100 \mathrm{~cm}$ has three resonant frequencies, $120 \mathrm{~Hz}, 200 \mathrm{~Hz}$ and $280 \mathrm{~Hz}$. If a node is formed at the end of the string, the speed of the transverse wave on this string is :
PhysicsWaves and SoundTS EAMCETTS EAMCET 2019 (04 May Shift 2)
Options:
  • A $60 \mathrm{~m} / \mathrm{s}$
  • B $80 \mathrm{~m} / \mathrm{s}$
  • C $100 \mathrm{~m} / \mathrm{s}$
  • D $120 \mathrm{~m} / \mathrm{s}$
Solution:
2184 Upvotes Verified Answer
The correct answer is: $80 \mathrm{~m} / \mathrm{s}$
Given, length of string, $l=100 \mathrm{~cm}=1 \mathrm{~m}$
Three resonant frequencies are given as $f_1=120 \mathrm{~Hz}, f_2=200 \mathrm{~Hz}, f_3=280 \mathrm{~Hz}$
$\because$ Fundamental frequency $\left(f_0\right)$ is equal to the greatest common factor of $f_1, f_2$ and $f_3$.

$\therefore f_0=$ greatest common factor of $120 \mathrm{~Hz}, 200 \mathrm{~Hz}$ and $280 \mathrm{~Hz}$ i.e., $f_0=40 \mathrm{~Hz}$
Now, we know that, speed of transverse wave is given by $v=2 l f_0=2 \times 1 \times 40=80 \mathrm{~m} / \mathrm{s}$.

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.