A wire of length 30cm , stretched between rigid supports, has it's...

Frequency of ${n}^{th}$ harmonic is $\dfrac{nv}{2l} = 400 ⇒ \dfrac{nv}{0 . 6} = 400 . . . \left(1\right)$

And frequency of ${\left(n + 1\right)}^{th}$ harmonic is $\dfrac{\left(n + 1\right)v}{2l} = 450 ⇒ \dfrac{\left(n + 1\right)v}{0 . 6} = 450 . . . \left(2\right)$, here, $v$ is speed of wave on string.

From equation \left(1\right)$, we have</p><p>$n = \dfrac{400 \times 0 . 6}{v}$</p><p>Putting in equation \left(2\right)$,

$⇒ \left[\dfrac{0 . 6 \times 400}{v} + 1\right]\dfrac{v}{0 . 6} = 450$

$⇒ \left[400 + \dfrac{v}{0 . 6}\right] = 450$

$⇒ v = 30m{s}^{- 1}$

Speed of wave on string is stated as $v = \sqrt{\dfrac{T}{\mu}} = 30$, here, $\mu$ is linear mass density.

So, $\mu = \dfrac{T}{{v}^{2}} = \dfrac{2700}{900} ⇒ \mu = 3kg{m}^{- 1}$