A string in a musical instrument is $50 \mathrm{~cm}$ long and its fundamental frequency is $800…

A string in a musical instrument is $50 \mathrm{~cm}$ long and its fundamental frequency is $800 \mathrm{~Hz}$. Keeping the tension applied to the string same, the length to produce sound note of fundamental frequency $1000 \mathrm{~Hz}$ will be.
  1. $10 \mathrm{~cm}$
  2. $20 \mathrm{~cm}$
  3. $60 \mathrm{~cm}$
  4. $40 \mathrm{~cm}$

Solution

For stretched string $\mathrm{v} \propto \frac{1}{l}$ $\therefore \frac{\mathrm{v}_1}{\mathrm{v}_2}=\frac{l_2}{l_1} \Rightarrow l_2=\frac{800}{1000} \times 50=40 \mathrm{~cm}$ /

Asked in: MHT CET 2022 (07 Aug Shift 1)

Practice more Waves and Sound questions on Aicharya