Question

Show that when a string fixed at its two ends vibrates in 1 loop, 2 loops, 3 loops, and 4 loops, the frequency are in the ratio 1:2:3:4.

Answer 1

Length for each loop = \(\frac{\lambda}{2}\)

Now,

L = \(\frac{n\lambda}{2}\) 

λ = \(\frac{2L}{n}\)   (1)

But v = vλ   or  λ = \(\frac{v}{u}\)

Putting in eqn, (1)

\(\frac{v}{u}=\frac{2L}{n}\)

v = \(\frac{n}{2L}\)u  

v = \(\frac{n}{2L}\sqrt{\frac{T}{\mu}}\)    [∵ u = \(\sqrt{\frac{T}{\mu}}\) ]

For n = 1, v₁ = \(\frac{1}{2L}\sqrt{\frac{T}{m}}\) = v₀

For n = 2, v₂ = \(\frac{1}{2L}{\sqrt{\frac{T}{m}}}\) = 2v₀

Therefore, u₁: u₂:u₃: u₄ = n₁: n₂: n₃:n₄

u₁: u₂:u₃: u₄ = 1: 2: 3: 4