Explain the formation os stationary waves by analytical method. Show the formation of stationary wave diagramatically.
A mass of 1 kg is hung from a steel wire of radius 0.5 mm and length 4 m. Calculate the extension produced. What should be the area of cross-section of the wire so that elastic limit is not exceeded? Change in radius is negligible ?
Given `g=9.8m//s^(2)`, Elastic limit of steel is `2.4xx10^(8)N//m^(2)`, Y for steel `(Y_("steel"))=20xx10^(10)N//m^(2),pi=3.142`).

Explanation

Formation of stationary waves :
Consider two simple harmonic progressive waves of equal amplitude and frequency propagating on a long uniform string in opposite directions.
If wave with frequency n and wavelength `lambda` is travelling along positive direction of X-axis then
`y_(1)=a sin 2pi(nt-(x)/(lambda))`
If wave with frequency n and wavelength `lambda` is travelling along negative direction of X-axis then
`y_(2)=a sin 2pi(nt+(x)/(lambda))`
The resultant displacement of stationary wave is given by the principle of superposition of waves.
`y=y_(1)+y_(2)`
`y=a sin 2pi(nt+(x)/(lambda))+a sin 2pi(nt-(x)/(lambda))`
`y=2a[sin(2pi nt)cos((2pi x)/(lambda))]`
`y=(2a " cos" (2pi x)/(lambda))sin (2pi nt)`
`y=A sin (2 pi nt)`
`y=A sin omega t " " (because omega =2pi n)`
where, ` A=2a "cos" (2pi x)/(lambda)`
(amplitude of resultant stationary wave)
Amplitude is periodic in space, hence we can see loops in case of transverse waves forming stationary wave on string.
This equation represents the resultant displacement of two simple harmonic progressive waves which is not moving. Thus, the waves are said to be stationary waves.

Numerical : Given : `M=1kg`
`r=0.5 mm =0.5 xx 10^(-3) m.`
`L=4m`
Y for steel `=20xx10^(10)N//m^(2)`
Elastic limit of steel `=2.4 xx 10^(8)N//m^(2)`
Now, Young's Modulus,
`Y=(FL)/(Al)`
`therefore l=(FL)/(pi r^(2)Y) " " [because F=mg]`
`l=(1 xx 9.4 xx 4)/(3.142 xx (0.5xx10^(-3))^(2)(20xx10^(10)))`
`=0.495 mm.`
Elastic limit of steel `=2.4 xx 10^(8)N//m^(2)`
If r' be the minimum radius of wire then,
Elastic limit `=(F)/(pi r'^(2))`
Elastic limit of steel = Maximum stress the wire can withstand
`(F)/(pi r'^(2))=2.4 xx 10^(8)`
`pi r^(2)=(F)/(2.4xx10^(8))`
`=(1xx9.8)/(2.4xx10^(8))`
`therefore` Area of cross section of wire (A')
`=4.083 xx 10^(-8)m. `
`=4.083xx10^(-5) mm.