Physics Circle home Student Login

Home›Chapters›Wave Motion PYQs›Stationary Waves & Vibrating Strings›Question 64

JEE Main 2023 (15 Apr, Shift 1) · Stationary Waves & Vibrating Strings

Stationary Waves & Vibrating Strings question from JEE Main 2023 (15 Apr, Shift 1)

Q64JEE Main 2023 · 15 APR · SHIFT 1Medium

Stationary Waves & Vibrating Strings

The fundamental frequency of vibration of a string stretched between two rigid supports is 50 Hz50\text{ Hz}. The mass of the string is 18 g18\text{ g} and its linear mass density is 20 g/m20\text{ g/m}. The speed of the transverse waves so produced in the string is ___________ m s−1\text{m s}^{-1}

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 90

  1. Length L=mμ=1820=0.9 mL = \dfrac{m}{\mu} = \dfrac{18}{20} = 0.9\text{ m}.
  2. Fundamental: n=v2L⇒v=2Ln=2×0.9×50n = \dfrac{v}{2L} \Rightarrow v = 2Ln = 2\times0.9\times50
  3. v=90 m/sv = 90\text{ m/s}
Report a mistake

More questions on Stationary Waves & Vibrating Strings