Question

A guitar string with a linear density of 2.0 g/m is stretched between supports that are 60 cm apart. The string is observed to form a standing wave with three antinodes when driven at a frequency of 420 Hz.

Part A
What is the frequency of the fifth harmonic of this string?
Part B
What is the tension in the string?

Answer 1

(a)
`f_(5)=700Hz`

(b)
`T=56.45N`

Step-by-step explanation:

For part (a)

The wave with three antinodes has m=3.Thus f₃=3f₁ and so the fundamental frequency of string is:

`f_(3)=3f_(1)\\f_(1)=(f_(3))/(3)\\ f_(1)=(420Hz)/(3)\\ f_(1)=140Hz`

Thus the frequency of fifth harmonic is

`f_(5)=5f_(1)\\f_(5)=5*140Hz\\f_(5)=700Hz`

For Part (b)

The speed of transverse wave on this string is related to the fundamental frequency by:

`f_(1)=(v)/(2l)\\ v=2lf_(1)\\v=2(0.6m)(140Hz)\\v=168m/s`

As the speed of wave is related to tension of string and its linear density by:

`v=\sqrt{(T)/(u) } \\T=v^(2)u\\T=(168m/s)^(2)(0.0020kg/m)\\T=56.45N`