A stretched string of length 1 m and mass 5 × 10-4 kg fixed at both ends is under tension of 20 N. it is plucked at a point situated at 25 cm from one end. The frequency of vibration of the string will be

Question

A stretched string of length 1 m and mass 5 × 10-4 kg fixed at both ends is under tension of 20 N. it is plucked at a point situated at 25 cm from one end. The frequency of vibration of the string will be (A) 125 Hz (B) 50 Hz (C) 200 Hz (D) 100 Hz

Tardigrade Admin · Accepted Answer

The frequency (n) of vibration of string is given by n=(p/2 l) √(T/m) where T is tension, and m the mass per unit length. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/biology/2627964c15fb9ee37e57c888c169ecc4-.png" /> Given, p=2, m=5 × 10-4 kg, T=20 N , l=1 m ∴ n=(2/2 × 1) √(20/5 × 10-4) =200 Hz

Q. A stretched string of length $1 m$ and mass $5 \times 1 0^{- 4} k g$ fixed at both ends is under tension of $20 N$ . it is plucked at a point situated at $25 c m$ from one end. The frequency of vibration of the string will be

125 Hz

50 Hz

200 Hz

100 Hz

The frequency $(n)$ of vibration of string is given by
$n = \frac{p}{2 l} \frac{T}{m}$
where $T$ is tension, and $m$ the mass per unit length.

Given, $p = 2, m = 5 \times 1 0^{- 4} k g$ ,
$T = 20 N, l = 1 m$
$∴ n = \frac{2}{2 \times 1} \frac{20}{5 \times 1 0 ^{- 4}}$
$= 200 Hz$

Q. A stretched string of length 1m and mass 5×10−4kg fixed at both ends is under tension of 20N. it is plucked at a point situated at 25cm from one end. The frequency of vibration of the string will be