#### Question

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

#### Solution

Let ‘l’ be the length of the wire which emits a fundamental note of frequency 256 Hz. When length = (l − 10) cm, fundamental frequency n = 320 Hz.

We know that the fundamental frequency n of a stretched string is given by

`n=1/(2l)sqrt("T"/"m")`

where ‘T’ is the tension and ‘m’ the linear density of the string.

When length = l, n = 256 Hz

i.e.

`256=1/(2l)sqrt("T"/"m")`......................(1)

When length = (l − 10) cm, n = 320 Hz

`320=1/(2(l-10))sqrt("T"/"m")`..........................................(2)

Dividing (1) by (2) gives

`256/320=(2(l-10))/(2l)`

`4/5=[(l-10)]/l`

**l 50 cm = 0.5 m**

Therefore, the original length of the wire is 50 cm.