The fundamental frequency of a sonometer wire is n. If its radius is doubled and its tension becomes half, the material of the wire remains same, the new fundamental frequency will be:

Question

The fundamental frequency of a sonometer wire is n. If its radius is doubled and its tension becomes half, the material of the wire remains same, the new fundamental frequency will be: (A)  n  (B)  (n/√2)  (C)  (n/2)  (D)  (n/2√2)

Tardigrade Admin · Accepted Answer

Frequency of sonometer wire is given by

n = \frac{1}{2 l} \frac{T}{m}

where m is mass of string per unit length, and is tension in the string.
Also,

m = π r^{2} d

r being radius of string and d is the density of material of string.
So,

n = \frac{1}{2 l} \frac{T}{π r ^{2} d}

or

n \propto \frac{T}{r}

or

\frac{n _{1}}{n _{2}} = \frac{T _{1}}{T _{2}} \times (\frac{r _{2}}{r _{1}})

Given,

r_{2} = 2 r_{1}, T_{2} = \frac{T _{1}}{2},

n_{1} = n

Hence,

\frac{n}{n _{2}} = 2 \times 2

or

n_{2} = \frac{n}{2 2}

Q. The fundamental frequency of a sonometer wire is n. If its radius is doubled and its tension becomes half, the material of the wire remains same, the new fundamental frequency will be:

$n$

$\frac{n}{2}$

$\frac{n}{2}$

$\frac{n}{2 2}$

Q. The fundamental frequency of a sonometer wire is n. If its radius is doubled and its tension becomes half, the material of the wire remains same, the new fundamental frequency will be:

n

2​n​

2n​

22​n​

$n$

$\frac{n}{2}$

$\frac{n}{2}$

$\frac{n}{2 2}$