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Waves question

2023 · 15 Apr · Shift 1 · Q64
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Waves question

2023 · 15 Apr · Shift 1 · Q64

JEE MainPhysicsWavesNumerical+4 / −1
The fundamental frequency of vibration of a string stretched between two rigid support is 50 Hz50 \mathrm{~Hz}50 Hz. The mass of the string is 18 g18 \mathrm{~g}18 g and its linear mass density is 20 g/m20 \mathrm{~g} / \mathrm{m}20 g/m. The speed of the transverse waves so produced in the string is ‾ms−1\underline{\hspace{2cm}}\mathrm{ms}^{-1}​ms−1
Numerical answer
View written solutionFree

Correct answer: 90

  1. Given data

    • Fundamental frequency: f1=50 Hzf_1 = 50\,\text{Hz}f1​=50Hz
    • Mass of string: m=18 g=0.018 kgm = 18\,\text{g} = 0.018\,\text{kg}m=18g=0.018kg
    • Linear mass density: μ=20 g/m=0.02 kg/m\mu = 20\,\text{g/m} = 0.02\,\text{kg/m}μ=20g/m=0.02kg/m

  2. Find the length of the string

    We use μ=mL\mu = \frac{m}{L}μ=Lm​ so, L=mμ=0.0180.02=0.9 mL = \frac{m}{\mu} = \frac{0.018}{0.02} = 0.9\,\text{m}L=μm​=0.020.018​=0.9m

  3. Use the formula for fundamental frequency of a stretched string

    For a string fixed at both ends, f1=v2Lf_1 = \frac{v}{2L}f1​=2Lv​ where vvv is the wave speed.

    Hence, v=2Lf1v = 2Lf_1v=2Lf1​

  4. Substitute the values

    v=2×0.9×50=90 m/sv = 2 \times 0.9 \times 50 = 90\,\text{m/s}v=2×0.9×50=90m/s

  5. Final answer

    The speed of transverse waves in the string is 90 m/s\boxed{90\,\text{m/s}}90m/s​

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