Question

A stretched wire emits a fundamental note of frequency 256Hz. Keeping the stretchin force constant and reducing the length of the wire by 10cm, frequency becomes 320Hz. Calculate original length of the wire.

Answer 1

If l is original length of wire, then
`256=(l)/(2l)sqrt((T)/(m))` and `320=(1)/(2(l-10))sqrt((T)/(m))`
Dividing, we get `(256)/(320)=(l-10)/(l)=(4)/(5)`
`5l-50=4l,l=50cm`