Q143 / 150 · MHT CET 2024 (10 May Shift 2) · Physics
Stationary wave is produced along the stretched string of length 80 cm . The resonant frequencies of string are $90 \mathrm{~Hz}, 150 \mathrm{~Hz}$ and 210 Hz . The speed of transverse wave in the string is
- A$45 \mathrm{~m} / \mathrm{s}$
- B$75 \mathrm{~m} / \mathrm{s}$
- Correct answerC$48 \mathrm{~m} / \mathrm{s}$
- D$80 \mathrm{~m} / \mathrm{s}$
Solution
$\mathrm{f}_1=90 \mathrm{~Hz}, \mathrm{f}_2=150 \mathrm{~Hz}, \mathrm{f}_3=210 \mathrm{~Hz}$
Fundamental frequency, $\mathrm{f}=30 \mathrm{~Hz}$ $\mathrm{f}_1=3 \mathrm{f}$
Length of string, $L=\frac{n \lambda}{2}$ $\lambda=\frac{2 \mathrm{~L}}{\mathrm{n}}=\frac{2 \times 80}{3 \times 100}=\frac{160}{300} \mathrm{~m}$ $\begin{aligned} & \text { Speed, } v=f \lambda \\ & v=90 \times \frac{160}{300}=48 \mathrm{~m} / \mathrm{s}\end{aligned}$
Fundamental frequency, $\mathrm{f}=30 \mathrm{~Hz}$ $\mathrm{f}_1=3 \mathrm{f}$
Length of string, $L=\frac{n \lambda}{2}$ $\lambda=\frac{2 \mathrm{~L}}{\mathrm{n}}=\frac{2 \times 80}{3 \times 100}=\frac{160}{300} \mathrm{~m}$ $\begin{aligned} & \text { Speed, } v=f \lambda \\ & v=90 \times \frac{160}{300}=48 \mathrm{~m} / \mathrm{s}\end{aligned}$
Hence, the correct answer is option: 'C'.