Question

Fundamental frequency of sonometer wire is n . If the v length , tension and diameter of wire are tripled the new fundamental frequency is?

Answer 1

Sonometer.
Radius = r. Density = d. Tension in the wire = T.

μ = mass/ length = d * pi* r^2
Velocity = sqrt(T/μ).

Fundamental tone Wavelength = lambda = 2* L.

Frequency n = 1/(2L) * sqrt (T/(pi *d*r^2))

If L, T and r are 3 times then:
New n = 1/(2*3*L)* sqrt(3*T/(pi*3d*9r^2))
= 1/9 * old n.

Frequency reduced to 1/9 times.

Answer 2

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 –

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

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

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

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

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