Question

Question #46523

Answer 1

`20cm`

Explanation:

Assuming that we are referring to the fundamental vibrational mode of the stretched string.

We know that in such a case the wavelength of the note is twice the length of the string.
`L=lambda/2` ......(1)
It can be also be shown that
Wave velocity `v=sqrt(T/rho)`
where `T` is tension in the string and `rho` is mass per unit length of wire.

Using the wave equation and `L` as length of string, we get
Frequency `f_1=v/lambda=sqrt(T/rho)/(2L)` .....(2)

Case 1. Given `256=sqrt(T/rho)/(2xx80)`
`=>sqrt(T/rho)=256(2xx80)` .....(3)

case 2. To find `1024=sqrt(T/rho)/(2xxL)`
Since it is similar wire under similar tension, using (3) we get
`1024=(256(2xx80))/(2xxL)`
`=>L=(256(2xx80))/(2xx1024)`
`=>L=20cm`