A travelling harmonic wave on a string is described by `y(x,t)=7.5sin(0.0050x +12t+pi//4)` (a) what are the displacement and velocity of oscillation of a point at `x=1cm, ` and `t=1s` ? Is this velocity equal to the velocity of wave propagation ?
(b) Locate the point of the string which have the same transverse displacement and velocity as `x=1cm` point at `t=2s, 5s and ` 11s.

Explanation

(a) The given harmonic wave is:
`y(x,t)=7"sin"(0.0050x+12t+(pi)/(4))`
For x=1 cm and t= 1s
`y(1,1)=7.5 "sin"(0.0050+12+(pi)/(4))`
`=7.5"sin"(12.0050+(pi)/(4))`
`=7.5 "sin"theta`
Where, `theta = 12.0050+(pi)/(4)=12.0050+(3.14)/(4)=12.79` rad
`=(180)/(3.14)xx12.79=732.81^(@)`
`therefore y(1,1)=7.5 "sin"(732.81^(@))`
`=7.5 "sin"(90xx8+12.81)=7.5 "sin"12.81^(@)`
`=7.5xx0.2217`
`=1.6629 ~~ 1.663 cm`
The velocity of the oscillation at a given point and time is given as:
`v=(d)/(dt)y(x,t)=(d)/(dt)[7.5 "sin"(0.0050x+12t+(pi)/(4))]`
`=7.5xx12"cos"(0.0050x+12t+(pi)/(4))`
At x=1cm and t=1 s:
`v=y(1,1)=90 "cos"(12.005+(pi)/(4))`
`=90"cos"(732.81^(@))=90"cos"(12.81^(@))`
`=90xx0.975=87.75 cm//s`
Now, the equation of a propagating wave is given by:
`y(x,t)=a"sin"(kx+wt+phi)`
Where,
`k=(2pi)/(lambda)`
`therefore lambda=(2pi)/(k)`
And `omega=2piv`
`therefore v=(omega)/(2pi)`
Speed , `v=vlambda=(omega)/(k)`
Where,
`omega=12` rad/s
`k=0.0050 m^(-1)`
`therefore v=(12)/(0.0050)=2400` cm/s
Hence, the velocity of the wave oscillation at x = 1 cm and t = 1 s is not equal to the velocity of the wave propagation.
(b) Propagation constant is related to wavelength as:
`k=(2pi)/(lambda)`
`therefore lambda=(2pi)/(k)=(2xx3.14)/(0.0050)`
`=1256 cm =12.56 m`
Therefore, all the points at distances `nlambda`
`(n+-1,+-2... "and so on")`, i.e. `+-12.56 m, +-25.12m,...` and so on for x=1 cm, will have the same displacement as the x=1 cm points at t=2 s, 5 s , and 11 s