Question

for a stationary wave set up in a string having both ends fixed what is the ratio of the fundamental frequency to the second harmonic?​

Answer 1

Given:

A stationary wave is set up with 2 ends fixed.

To find:

The ratio of the fundamental frequency to the second harmonic?

Calculation:

Fundamental frequency:

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

$\implies \: \lambda = 2l$

$\implies \: f_{1} = \dfrac{v}{ \lambda}$

$\boxed{\implies \: f_{1} = \dfrac{v}{2l} }$

For 2nd harmonic:

$\therefore \: l = \dfrac{ \lambda}{2} + \dfrac{ \lambda}{2}$

$\implies \: l = \lambda$

$\implies \: f_{2} = \dfrac{v}{ \lambda}$

$\boxed{ \implies \: f_{2} = \dfrac{v}{l} }$

Required ratio:

$\boxed{ \bf \therefore \: f_{1} : f_{2} = 1 : 2}$

Answer 2

Explanation:

Given:

A stationary wave is set up with 2 ends fixed.

To find:

The ratio of the fundamental frequency to the second harmonic?

Calculation:

Fundamental frequency:

\therefore \: l = \dfrac{ \lambda}{2} ∴l=

2

λ

\implies \: \lambda = 2l⟹λ=2l

\implies \: f_{1} = \dfrac{v}{ \lambda} ⟹f

1

=

λ

v

\boxed{\implies \: f_{1} = \dfrac{v}{2l} }

⟹f

1

=

2l

v

For 2nd harmonic:

\therefore \: l = \dfrac{ \lambda}{2} + \dfrac{ \lambda}{2} ∴l=

2

λ

+

2

λ

\implies \: l = \lambda⟹l=λ

\implies \: f_{2} = \dfrac{v}{ \lambda} ⟹f

2

=

λ

v

\boxed{ \implies \: f_{2} = \dfrac{v}{l} }

⟹f

2

=

l

v

Required ratio:

\boxed{ \bf \therefore \: f_{1} : f_{2} = 1 : 2}

∴f

1

:f

2

=1:2