Q141 / 160 · TS EAMCET 2019 (03 May Shift 1) · Physics
Standing waves are produced in a string 16 $\mathrm{m}$ long. If there are 9 nodes between the two fixed ends of the string and the speed of the wave is $32 \mathrm{~m} / \mathrm{s}$, what is the frequency of the wave?
- A$5 \mathrm{~Hz}$
- Correct answerB$10 \mathrm{~Hz}$
- C$30 \mathrm{~Hz}$
- D$20 \mathrm{~Hz}$
Solution
Given, length of spring, $l=16 \mathrm{~m}$
speed of wave, $v=32 \mathrm{~m} / \mathrm{s}$
and total number of nodes between the two fixed ends of the string $=9$
$\therefore$ Total number of segments of vibrations between fixed ends, $p=9+1=10$
$\therefore$ frequency of the wave, $f=\frac{p}{2 L} \times v$
$$
=\frac{10}{2 \times 16} \times 32=10 \mathrm{~Hz}
$$
Hence, the correct answer is option: 'B'.