Question

What is the natural frequency of a sonometer wire under tension ?

​

Answer 1

${\huge{\bold{Heya \:Mate}}}$

${\bold{\underline{Here's \:The\:Answer}}}$

→ What is The natural frequency of a sonometer wire under some tension ?

=> The natural frequency of a sonometer wire under tension is given by the formula :

$n = \frac{1}{2l} \sqrt{ \frac{t}{m} }$

Where

↪ T is the tension of the wire

↪m is the mass per unit length of the wire

↪ l is the length of the wire between the bridges

${\mathbb{\bold{Hope\:this\:Helps!!}}}$

__________________________________________________

Answer 2

Explanation:

ANSWER :–

new frequency – $$\begin{lgathered}{ \boxed { \bold { {n}_{1} = \frac{n}{4} }}} \\\end{lgathered}$$

EXPLANATION :–

GIVEN :–

• The fundamental frequency of a sonometer wire is n.

• The length and diameter of the wire are doubled keeping the tension same.

TO FIND :–

New fundamental frequency = ?

SOLUTION :–

• We know that frequency –

$$\begin{lgathered}\\ { \boxed { \bold { n = \frac{1}{2l} \sqrt{ \frac{T}{m} }}}} \\\end{lgathered}$$

• Mass per unit Length :–

m = [ρΠr²(l)]/l

m = ρΠr²

• So that , new equation –

$$\begin{lgathered}\\ { \boxed { \bold { n = \frac{1}{2(2l)} \sqrt{ \frac{T}{ \rho \pi {(2r)}^{2}}}}}} \\\end{lgathered}$$

• Now According to the question –

Length (l) => 2l and radius (r) => 2r

• New fundamental frequency –

$$\begin{lgathered}\\ { \bold { {n}_{1} = \frac{1}{2(2l)} \sqrt{ \frac{T}{ \rho \pi {(2r)}^{2}} }}} \\\end{lgathered}$$

$$\begin{lgathered}\\ { { \bold { {n}_{1} = \frac{1}{4l} \sqrt{ \frac{T}{4 \rho \pi {r}^{2}} }}}} \\\end{lgathered}$$

$$\begin{lgathered}\\ { { \bold { {n}_{1} = \frac{1}{2(4l)} \sqrt{ \frac{T}{ \rho \pi {r}^{2}} }}}} \\\end{lgathered}$$

$$\begin{lgathered}\\ { { \bold { {n}_{1} = \frac{1}{4} {n} }}} \\\end{lgathered}$$

$$\begin{lgathered}\\ { { \bold { {n}_{1} = \frac{n}{4} }}} \\\end{lgathered}$$