Question

Two pieces of steel wire with identical cross sections have lengths of L and 2L. The wires are each fixed at both ends and stretched so that the tension in the longer wire is four times greater than in the shorter wire. If the fundamental frequency in the shorter wire is 60 Hz, what is the frequency of the second harmonic in the longer wire?

Answer 1

Answer:

Explanation:

Expression for fundamental  frequency of tone produced in a wire under tension of T and length L is given as follows

$f=\frac{1}{2L} \times \sqrt{\frac{T}{ m} }$

m is mass per unit length .

We shall apply this formula for given wires .

For shorter wire

$60 =\frac{1}{2L} \times \sqrt{\frac{T}{ m} }$

For longer wire for second harmonic

length of wire is 2L , tension is 4T ,

$f =\frac{2}{4L} \times \sqrt{\frac{4T}{ m} }$

$f =\frac{2\times 2}{4L} \times \sqrt{\frac{T}{ m} }$

f = 2 x 60 = 120 Hz .