Q111 / 150 · MHT CET 2023 (11 May Shift 1) · Physics
A string fixed at both the ends forms standing wave with node separation of $5 \mathrm{~cm}$. If the velocity of the wave on the string is $2 \mathrm{~m} / \mathrm{s}$, then the frequency of vibration of the string is
- A$0.2 \mathrm{~Hz}$
- B$10 \mathrm{~Hz}$
- Correct answerC$20 \mathrm{~Hz}$
- D$40 \mathrm{~Hz}$
Solution
Separation between consecutive nodes, $\frac{\lambda}{2}=5 \mathrm{~cm}$
$$
\therefore \quad \lambda=10 \mathrm{~cm}=0.1 \mathrm{~m}
$$
The frequency of vibration is given as:
$$
\mathrm{n}=\frac{\mathrm{v}}{\lambda}=\frac{2}{0.1}=20 \mathrm{~Hz}
$$
Hence, the correct answer is option: 'C'.