Analytical treatment of stationary waves : Consider two identical simple harmonic progressive waves of equal amplitude and frequency travelling along X-axis in opposite direction. They are givne by , the wave along positive X-axis.
`y_(1) = a sin 2 pi( nt-(x)/(lambda))" "...(i)`
The wave along negative direction of x-axis is given by :
`y_(2) = a sin 2pi (nt+(x)/(lambda))" "..(2)`
By principle of superposition of waves, the resultant displacement of stationary wave is given by
`y=y_(2) +y_(1)`
` y= a sin 2 pi (nt+(x)/(lambda))+a sin 2 pi(nt-(x)/(lambda))`
`:. y= 2a "sin" (2pi)/(2) (nt-(x)/(lambda)+nt+(x)/(lambda)) "cos"(2pi)/(2)(nt-(x)/(lambda)-nt-(x)/(lambda))`
[Using `sin C+ sin D= 2 sin ((C+D)/(2)) cos ((C-D)/(2))]`
`y= 2a sin (2 pi nt)* cos 2 pi (-(x)/(lambda))`
`:. y=2a[ sin (2pi nt)* cos ((2pi x)/(lambda))] [ :. cos (-theta)= cos theta]`
`y=[ 2 a cos ((2pix)/(lambda))] sin 2 pi nt `
`y= A sin (2pi nt)`
Where, `A=2 a "cos" (2pix)/(lambda)`
A is the amplitude of the result wave is a simple harmonic motion having same period but new amplitude. The absence of the (x) in the sine function shows that the resultant do not move forward or backward. Such waves are called stationary waves.
Condition for Antinode : At antinodes,
`A= pm2a`
`:. 2a "cos" (2pix)/(lambda)=pm2a` ltbrlt `rArr "cos"(2pirx)/(lambda)= pm1`
`rArr (2pix)/(lambda)0,pi, 2pi, ..... p pi`
`rArr (2pix)/(lambda)=p pi`
`x=(lambdap)/(2)` where, `p=0,1,2....`
When `(2pix)/(lambda)=0` then `x=0`
When `(2pir)/(lambda)=pi`, then `x=(lambda)/(2)`
When `(2pix)/(lambda)=2pi`, then `x=lambda`
`:. x=0,(lambda)/(2),.....`
`:.` Distance between two consecutive antinodes
`=x_(1)-x_(0)=lambda//2`
`=x_(2)-x_(1)=(lambda)/(2)` and so on.
Thus, distance between two sucessive antinods is `lambda//2`
Condition for nodes : At nodes,
`:. 1a cos ((2pi x)/(lambda))=0`
`rArr cos ((2pix)/(lambda))=0`
`rArr (2pix)/(lambda)=(pi)/(2),(3pi)/(2),(5pi)/(2)......`
`rArr x=(((2p-1)lambda)/(4))`
where `p=1,2,3,...`
When `(2pix)/(lambda)=(pi)/(2)` then `x=(lambda)/(4)`
When `(2pir)/(lambda)=(3pi lambda)/(2)` then, `x=(3lambda)/(4)`
When `(2pix)/(lambda)=(5pi)/(2)`, then `x=(5lambda)/(4)`
`:. x=(lambda)/(4),(3lambda)/(4),(5lambda)/(4).....`
The particles at these points vibrates with minimum or zero amplitudel. The distance. between any two successive nodes is `(lambda)/(2)`
Hence, the nodes and antinodes are equally spaced in stationary waves.
Numerical
Given `h=0.4m, k_(0) =0.5 m`
`k_(0)^(2)=k_(c)^(2)+r^(2)`
`k_(0)^(2)=k_(c)^(2)+r^(2)`
`:. (0.5)^(2)+k_(c)^(2)+(0.4)^(2)`
`rArr 0.25=k_(c)^(2)+0.16`
`rArr 0.25-0.16=k_(c)^(2)`
`rArr 0.9 =k_(c)^(2)`
`k_(c)=0.3m`
Hence radius of gyration about a parallel axis passing through its centre of mass is `0.3m`
