A wire having a linear mass density `10^(-3) kg//m` is stretched between two rigid supports with a tension of `90 N`. This wire resontes at a frequency of `350 Hz`. The next higher frequency at which the same wire resonates is `420 Hz`. Find the length of the wire.


Share with your friends
Call

Suppose the wire vibrates at `350 Hz` in its nth harmonic and at `420 Hz` in its `(n + 1)` th harmonic.
`350s^(-1) = (n)/(2 L) sqrt((F)/(mu)) ….(i)`
and `420s^(-1) = ((n + 1))/(2l.) sqrt((F)/(mu)) …..(ii)`
This gives `(420)/(350) = (n + 1)/(n)` or, `n = 5`.
Putting the value in `(i)`
`350 = (5)/(2l)sqrt((90)/(10^(-3))) rArr 350 = (5)/(2l) xx 300 rArr l = (1500)/(700) = (15)/(7)m = 2.1 m`

Talk Doctor Online in Bissoy App