If n_{1}, n_{2} and n_{3} are the fundamental | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The fundamental frequency of a stretched string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$.Since the tension $T$ and mass per unit length

Vedclass · Accepted Answer

$\frac{1}{n} = \frac{1}{n_{1}} + \frac{1}{n_{2}} + \frac{1}{n_{3}}$

Vedclass · Answer

(C) The fundamental frequency of a stretched string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$.Since the tension $T$ and mass per unit length $\mu$ are constant for the string,we have $n \propto \frac{1}{l}$,which implies $l = \frac{k}{n}$,where $k$ is a constant.When the string is divided into three segments of lengths $l_{1}, l_{2}, l_{3}$ with fundamental frequencies $n_{1}, n_{2}, n_{3}$ respectively,we have $l_{1} = \frac{k}{n_{1}}$,$l_{2} = \frac{k}{n_{2}}$,and $l_{3} = \frac{k}{n_{3}}$.The total length of the string is $l = l_{1} + l_{2} + l_{3}$.Substituting the expressions for length in terms of frequency,we get $\frac{k}{n} = \frac{k}{n_{1}} + \frac{k}{n_{2}} + \frac{k}{n_{3}}$.Dividing by $k$,we obtain the relation $\frac{1}{n} = \frac{1}{n_{1}} + \frac{1}{n_{2}} + \frac{1}{n_{3}}$.

If $n_{1}, n_{2}$ and $n_{3}$ are the fundamental frequencies of three segments into which a string is divided,then the original fundamental frequency $n$ of the string is given by

Similar Questions

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 $Hz$)

If the frequency of vibrations of a string is to be increased by a factor of two,then the tension in the string must be made:

$A$ wire of length $L$ and linear density $m$ is stretched between two rigid supports with tension $T$. It is observed that the wire resonates in the $P^{th}$ harmonic at a frequency of $320 \ Hz$ and resonates again at the next higher frequency of $400 \ Hz$. The value of $p$ is:

$A$ string of length $1 \, m$ fixed at both ends is vibrating in the $3^{rd}$ overtone. The tension in the string is $200 \, N$ and the linear mass density is $5 \, g/m$. The frequency of these vibrations is ..... $Hz$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module