A stretched string is 1 m long. Its mass per unit length is 0.5 g / m. It is stretched with a force of 20 N. It plucked at a distance of 25 cm from one end. The frequency of note emitted by it will be :

Question

A stretched string is 1 m long. Its mass per unit length is 0.5 g / m. It is stretched with a force of 20 N. It plucked at a distance of 25 cm from one end. The frequency of note emitted by it will be : (A) 400 Hz (B) 300 Hz (C) 200 Hz (D) 100 Hz

Tardigrade Admin · Accepted Answer

Using relation f=(p/2 l) √(T/m) (Since, string is plucked at 25 cm hence, it will vibrate in two segments p=2 ) f=(2/2 × 1) √(20/0.5 × 10-3) =200 Hz

Q. A stretched string is $1 m$ long. Its mass per unit length is $0.5 g / m$ . It is stretched with a force of $20 N$ . It plucked at a distance of $25 c m$ from one end. The frequency of note emitted by it will be :

400 Hz

300 Hz

200 Hz

100 Hz

Using relation $f = \frac{p}{2 l} \frac{T}{m}$
(Since, string is plucked at $25 c m$ hence, it will vibrate in two segments $p = 2$ ) $f = \frac{2}{2 \times 1} \frac{20}{0.5 \times 1 0 ^{- 3}}$
$= 200 Hz$

Q. A stretched string is 1m long. Its mass per unit length is 0.5g/m. It is stretched with a force of 20N. It plucked at a distance of 25cm from one end. The frequency of note emitted by it will be :