Question

The fundamental frequency of a sonometer wire is n. If the length and diameter of the wire are doubled keeping the tension same, then the new fundamental frequency is pls

Answer 1

ANSWER :–

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

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 –

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

• Mass per unit Length :–

m = [ρΠr²(l)]/l

m = ρΠr²

• So that , new equation –

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

• Now According to the question –

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

• New fundamental frequency –

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

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

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

$\\ { { \bold { {n}_{1} = \frac{1}{4} {n} }}}$

$\\ { { \bold { {n}_{1} = \frac{n}{4} }}}$