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 is 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 is plucked at a distance of 25 cm from one end. The frequency of note emitted by it will be: (A) 400 m (B) 300 Hz (C) 200 Hz (D) 100 Hz

Tardigrade Admin · Accepted Answer

As stretched string of 1 m long is plucked at a distance of 25 cm from end, so there will be 2 loops. Frequency, n=(p/2 l) √(T/m) Given, p=2, l=1 m, T=20 N, m=0.5 × 10-3 kg / m ∴ n=(2/2 × 1) √(20/0.5 × 10-3) ⇒ n=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 is plucked at a distance of 25cm from one end. The frequency of note emitted by it will be:

400 m

300 Hz

200 Hz

100 Hz

As stretched string of 1m long is plucked at a distance of 25cm from end, so there will be 2 loops. Frequency, n=2lp​mT​​ Given, p=2,l=1m,T=20N, m=0.5×10−3kg/m ∴n=2×12​0.5×10−320​​ ⇒n=200Hz

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 is plucked at a distance of $25 c m$ from one end. The frequency of note emitted by it will be: