A sonometer wire is vibrating in resonance


A sonometer wire is vibrating in resonance with a tuning fork. Keeping the tension applied same, the length of the wire is doubled. Under what conditions would the tuning fork still be is resonance with the wire?


The length of the wire used in a sonometer is twice when it vibrates. Therefore, if the tuning fork resonates at L, the sonometer resonates at 2L. The following equation is used to express the frequency of the sonometer

$f=\frac{n}{2 L} \sqrt{\frac{T}{\mu}}=\frac{n v}{2 L}$

The velocity of the wave from the sonometer is constant. The resonance between the tuning fork and the wire will remain even after the change in the length of the wire.


n/L = constant

Which also means that,

n1/L1 = n2/L2

n1/L1 = n2/2L2

n2 = 2n1

Therefore, when the wire length is doubled, the resonance also gets doubled.

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