Question

A thin inextensible string of length 1 meter is stretched by a weight of 1 kg. What should be the new weight so that the fundamental frequency of the string is doubled?

Answer 1

Given:

A thin inextensible string of length 1 meter is stretched by a weight of 1 kg.

To find:

New weight so that the fundamental frequency of the string is doubled?

Calculation:

For $\rm fun da m ental$ frequency , we can say:

$l = \dfrac{ \lambda}{2}$

$\implies \lambda = 2l$

Now, we can say:

$\implies f = \dfrac{v}{ \lambda}$

$\implies f = \dfrac{ \sqrt{ \dfrac{T}{ \mu} } }{ \lambda}$

$\implies f = \dfrac{ \sqrt{ \dfrac{mg}{ \mu} } }{2l}$

For constant $\mu$ and $\lambda$ (i.e. 2l), we can say:

$\implies \: f \propto \: \sqrt{m}$

• In order to double the frequency, the mass should be increased by 4 times.

So, new mass = 4 × 1 = 4 kg.

Answer 2

Explanation:

Given:

A thin inextensible string of length 1 meter is stretched by a weight of 1 kg.

To find:

New weight so that the fundamental frequency of the string is doubled?

Calculation:

For \rm fun da m entalfundamental frequency , we can say:

l = \dfrac{ \lambda}{2} l=

2

λ

\implies \lambda = 2l⟹λ=2l

Now, we can say:

\implies f = \dfrac{v}{ \lambda} ⟹f=

λ

v

\implies f = \dfrac{ \sqrt{ \dfrac{T}{ \mu} } }{ \lambda} ⟹f=

λ

μ

T

\implies f = \dfrac{ \sqrt{ \dfrac{mg}{ \mu} } }{2l} ⟹f=

2l

μ

mg

For constant \muμ and \lambdaλ (i.e. 2l), we can say:

\implies \: f \propto \: \sqrt{m} ⟹f∝

m

In order to double the frequency, the mass should be increased by 4 times.

So, new mass = 4 × 1 = 4 kg.