If the tension and diameter of a sonometer wire of fundamental frequency n is doubled and density is halved then its fundamental frequency will become

Question

If the tension and diameter of a sonometer wire of fundamental frequency n is doubled and density is halved then its fundamental frequency will become (A) (n/4) (B) √2n (C) n (D) (n/√2)

Tardigrade Admin · Accepted Answer

n=(1/2 l) √(T/π r2 ρ) ρ1 prime=(ρ/2) ; T prime=2 T and D prime=2 D or r prime=2 r n prime=(1/2 l) √(2 T/π(2 r)2 (ρ)2) after solving, n prime=(1/2 l) √(T/π r2 l)=n.

Q. If the tension and diameter of a sonometer wire of fundamental frequency $n$ is doubled and density is halved then its fundamental frequency will become

$\frac{n}{4}$

$2 n$

$n$

$\frac{n}{2}$

$n = \frac{1}{2 l} \frac{T}{π r ^{2} ρ}$
$ρ_{1}^{'} = \frac{ρ}{2}; T^{'} = 2 T$
and $D^{'} = 2 D$
or $r^{'} = 2 r$
$n^{'} = \frac{1}{2 l} \frac{2 T}{π ( 2 r ) ^{2} \frac{ρ}{2}}$
after solving, $n^{'} = \frac{1}{2 l} \frac{T}{π r ^{2} l} = n$ .

Q. If the tension and diameter of a sonometer wire of fundamental frequency n is doubled and density is halved then its fundamental frequency will become

4n​

2​n

n

2​n​

n=2l1​πr2ρT​​ ρ1′​=2ρ​;T′=2T and D′=2D or r′=2r n′=2l1​π(2r)22ρ​2T​​ after solving, n′=2l1​πr2lT​​=n.

Q. If the tension and diameter of a sonometer wire of fundamental frequency $n$ is doubled and density is halved then its fundamental frequency will become

$\frac{n}{4}$

$2 n$

$n$

$\frac{n}{2}$

$n = \frac{1}{2 l} \frac{T}{π r ^{2} ρ}$
$ρ_{1}^{'} = \frac{ρ}{2}; T^{'} = 2 T$
and $D^{'} = 2 D$
or $r^{'} = 2 r$
$n^{'} = \frac{1}{2 l} \frac{2 T}{π ( 2 r ) ^{2} \frac{ρ}{2}}$
after solving, $n^{'} = \frac{1}{2 l} \frac{T}{π r ^{2} l} = n$ .