$A$ string in a musical instrument is $50 \,cm$ long and its fundamental frequency is $800 \,Hz$. Keeping the tension applied to the string same, the change in the length to produce a sound note of fundamental frequency $1000 \,Hz$ will be: (in $\,cm$)

  • A
    $10$
  • B
    $20$
  • C
    $60$
  • D
    $40$

Explore More

Similar Questions

$A$ sonometer wire is vibrating in the third overtone. There are:

$A$ string fixed at both ends vibrates in three loops. The wavelength is $10 \, cm$. The length of the string is .... $cm$.

$A$ stretched string resonates with a tuning fork of frequency $512 \; Hz$ when the length of the string is $0.5 \; m$. The length of the string required to vibrate resonantly with a tuning fork of frequency $256 \; Hz$ would be .......... $m$.

In order to double the frequency of the fundamental note emitted by a stretched string, the length is reduced to $\frac{3}{4}$ of the original length and the tension is changed. The factor by which the tension is to be changed is

In a Sonometer experiment,the frequency of the tuning fork used is $288 \ Hz$. Harmonics will '$NOT$' be produced at the frequency: (in $Hz$)

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo