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

The given harmonic wave is
`y(x, t) = 7.5 sin (0.0050 x + 12 t + (pi)/(4))" "...(i)`
At x = 1 cm, t = 1s
`y(1, 1) = 7.5 sin (0.0050 xx 1 + 12 xx 1 + (pi)/(4))`
`= 7.5 sin (12.005 + (pi)/(4))`
`= 7.5 sin (12.005 + (3.14)/(4))`
`= 7.5 sin (12.79)`
`= 7.5 sin (theta)`
`theta = 12.79` rad
`= 12.79 xx (180)/(pi) = (12.79 xx 180)/(3.14) = 733.1^(@)`
`y(1,1) = 7.5 sin (733.1^(@))`
`= 7.5 sin (720 + 13.1)`
`= 7.5 sin 13.1^(@)`
`= 7.5 xx 0.226`
= 1.69 cm
The velocity of oscillation,
`V = 7.5 xx 120 cos (0.005 + 12 + (pi)/(4))`
At x = 1 cm, t = 1 s
`V = 90 cos (733.1^(@))`
`= 90 cos (720 + 13.1)`
`= 90 cos (13.1^(@))`
`= 90 xx 0.9739`
= 87.65 cm/s
Comparing equation (i) with this standard equation
`y(x,t) = r sin ((2pi)/(lambda) (upsilon t + x) + phi_(0))` we get
`r = 7.5 cm, (2pi)/(lambda) upsilon = 2pi v = 12`
`rArr " "v = (12)/(2pi) = (6)/(pi)`
`(2pi)/(lambda) = 0.005`
`lambda = (2pi)/(0.005) = 1256` cm
Velocity of wave propagation
`= V = v lambda = (6)/(pi) xx 1256 = 2400` cm/s
Thus velocity of wave propagation is not equal to velocity at x = 1 cm and t = 1 s
(b) All the points which are at a distance of `+- lambda, +- 2 lambda, +- 3 lambda`, will have the same velocity and displacement as at x = 1 cm
As `lambda = 12.56` m
therefore, the points at distance `+- 12.56, +- 25.2 m, +- 37.8` m will have the same displacement and velocity as x = 1 cm at points t = 2s, 5s and 11s.