Answer to Question #163193 in Optics for Keleko Pierre Michelle

Question

A 2.00m long wire having a mass of 0.100kg is fixed at both ends. The tension in the wire is maintained at 20.0N 

1. What are the frequencies of the first three allowed modes of vibration? 
2. If a node is observed at a point 0,400m from one end, in what mode with what frequency is it vibrating? 

Answer 1

Remember that the harmonics of a vibrating string have frequencies that are related by integer multiples of the fundamental frequency of standing wave. For second part first we find numbers of waves present in the string and finally using that wave length we find frequency of string.

From the data we can find the linear density of wire and it is used to find the fundamental frequency and next three allowed modes.

Given,

Length of wire $(L)=2.00m$

Mass of wire $(m)=0.100kg$

Liner density of the wire is

$\mu=\dfrac{m}{L}$

$= \dfrac{0.100kg}{2.00m}=0.05kg/m$

Tension in the wire $(T)=20.0N$

(a)Fundamental frequency of the wire is

$f_1=\dfrac{1}{2L} \sqrt{\dfrac{T}{\mu}}$

$=\dfrac{1}{2(2.00m)}\sqrt{\dfrac{20.0N}{0.05kg/m}}$

$= 5 Hz$

The frequencies of the remaining normal modes are integral multiples of the fundamental frequency.

Therefore the frequency of the second allowed mode of vibration is

$f_2=2f_1$

$=2(5Hz)$

$=$ 10Hz

The frequency of the third allowed mode of vibration is

$f_3=3f_1$

$=3(5Hz)$

$=$ 15 Hz

(b)The distance between the two nodes is

$\dfrac{\lambda}{2}= 0.400m$

So the wavelength is

$\lambda=0.800m$

Number of waves in the length is

$n=\dfrac{L}{\lambda}$

$=\dfrac{2.0m}{0.800m}=2.5$ waves

Numbers of modes present in the waves are $2.5\times 2=5$

Frequency of the wave is

$f=\dfrac{v}{\lambda}$

$=\dfrac{1}{\lambda}\sqrt{\dfrac{T}{\mu}}$

$=\dfrac{1}{(0.800m)}\sqrt{\dfrac{20.0N}{0.05kg/m}}$

= 25 Hz