A travelling harmonic wave on a string is described by `y(x,t) = 7.5 sin (0.005 x + 12t+ pi//4)` (a) what are the displacement and velocity of oscillation of a point at x=1 cm, and t=1s? Is this velocity equal to the velocity of wave propagation?
(b) Locate the points of the string which have the same transverse displacements and velocity as the x=1 cm point at t=2s, 5s and 11s

Explanation

`(i) y =7.5 sin (0.0050 x +12 t + (pi)/(4)) cm …(1)`
`y =a sin (kx + omega t + phi)" "…(2)`
Comparing (1) and (2),
we get `a = 7.5 cm" "…(3)`
`k = (2pi)/(lamda) = 0.005 (rad)/(cm) " "…(4)`
`omega =2pi f =12 (rad )/(s)" "...(5)`
`phi = (pi)/(4) rad " "...(6)`
Now, required displacemetn is,
`y =7.5 sin {(0.005 xx 1 ) + (12xx 1) + (pi)/(4)}`
`therefore y =7.5 sin (12.005 +(pi)/(4))`
`= 7.5 sin ((48.02 +3.14)/(4))`
`= 7.5 sin (12.79)`
`= 7.5 sin (12 .79 xx (180)/(3.14)) ^(@)`
`=7.5 sin (733^(@))`
`= 7.5 sin (720^(@) + 13^(@))`
`= 7.5 sin (13 ^(@))`
`= (7.5) (0.2250)`
`therefore =1.6875 cm`
Now velocity of oscilaation of above particle,
`v _(p) = (dy)/(dt) = (d)/(dt) {7.5 sin (0.005 x + 12 t + (pi)/(4))}`
`therefore v _(p) = 7.5 cos (0.005 x + 12 t + (pi)/(4)) {0+(12 xx 1) + 0}`
`=7.5 xx 12 cos { (0.005 xx 1 ) + (12 xx 1 ) + (pi)/(4) }`
`= 90 cos (13 ^(@))` (As simplified above)
`=(90) (0.9744)`
`v _(p) = 87.696 cm//s`
Now velocity of wave propagation is,
`v = (omega )/(k) = (12)/(0.005) = 2400 (cm)/(s)`
`impliesv ne v _(p)`
Moreover, directions of `vecv and vec(v_(p))` are also not same av `vecv bot vec(v _(p)) .` Thus `vecv ne vec(v _(p))`
Here,
`k = (2pi)/(lamda) = 0.005 (rad)/(cm) implies lamda = (2 xx 3.14)/(0.005) = 1256cm`
`implies` Along the length of string, particles situated at distances `lamda, 2 lamda, 3 lamda, 4 lamda, ...` will have same displacement `y =1.6875 cm` as that of particle at `x=1 cm ` and at `t=1s.`
(ii) Now, we have `x =1 cm and t = 5s.` Hence,
`y = 7.5 sin { (0.005 xx 1 ) + (12 xx 5) + (pi)/(4)}`
`= 7.5 sin [ 60.005 + (pi)/(4) ]`
`=7.5 sin ((240 .02 + 3.14)/(4))`
`=7.5 sin (60.79)`
`= 7.5 sin (60.79 xx (180)/(3.14) ) ^(@)`
`=7.5 sin (3485^(@)) =7.5 sin (9.68xx360)^(@)`
`therefore y =7.5 sin {(360^(@) xx 9) + (360 ^(@) xx 0.68)}`
`=7.5 sin {360^(@)xx 06.8)`
`=7.5 sin (245^(@))`
`=7.5 sin (180^(@) + 65 ^(@))`
`=7.5 {- sin 65 ^(@)}`
`=7.5 (-0.9063)`
`y =- 6.797 cm`
Along the length of string, particles situated at `lamda, 2lamda, 3lamd, 4lamda, ...` will have same displacement `y =-6.797cm` and as that of a particle at `x =1 cm at t =5s.`
(iii) Now, we have `x =1 cm and t = 11s.` Hence,
`y = 7.5 sin {(0.005xx 1) + (12 xx 1) + (pi)/(4) }`
`=7.5 sin (132.005 + (pi)/(4))`
`=7.5 sin ((528 .02 + 3.14)/(4))`
`= 7.5 sin (132.79)`
`=7.5 sin (132 .79 xx (180)/(3.14))^(@)`
`=7.5 sin (7612)^(@)`
`=7.5 sin (360^(@) xx 21.144 )`
`= 7.5 sin {(360 ^(@)xx 21) + (360 xx 0.144)}`
`= 7.5 sin (360^(@) xx 0.144)`
`=7.5 sin (52 ^(@))`
`=7.5 (0.7880)`
`y = 5.91 cm`
Along the length of the string, particles situated at `lamda, 2lamda, 3 lamda, 4 lamda, ...` will have same displacement `y=5.91cm` as that of a particle at `x=1 cm` at `t =11s.`