Ncert Exercises & Solutions

A travelling harmonic wave on a string is described by $y (x, t) = 7.5 \sin (0.0050 x + 12 t + \frac{π}{4})$ .

(a) what are the displacement and velocity of oscillation of a point at x = 1 cm, and t = 1 sec? 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 that of point at x = 1 cm and t = 2 s, 5 s and 11 s.

The given harmonic wave is:

$y (x, t) = 7.5 \sin (0.0050 x 12 t + \frac{π}{4})$
$For x = 1 cm and t = 1 s,$
$\begin{aligned} y (1, 1) & = 7 .5 \sin (0 .0050 + 12 + \frac{π}{4}) \end{aligned}$
$= 7 .5 \sin (12 .0050 + \frac{π}{4})$
$= 7 .5 sinθ$
$where, θ = 12 .0050 + \frac{π}{4} = 12 .0050 + \frac{3 .14}{4} = 12 .79 rad$
$= \frac{180}{3 .14} \times 12 .79 = 732 {.81}^{\circ}$
$\begin{aligned} ∴ y (1, 1) & = 7 .5 \sin (732 {.81}^{\circ}) \\ = 7 .5 \sin (90 \times 8 + 12 {.81}^{\circ}) = 7 .5 \sin 12 {.81}^{\circ} \\ = 7 .5 \times 0 .2217 \\ = 1 .6629 \approx 1 .663 cm \end{aligned}$
$The velocity of the oscillation :$
$\begin{aligned} v = \frac{d}{dt} y (x, t) & = \frac{d}{dt} [7 .5 \sin (0 .0050 x + 12 t + \frac{π}{4})] \\ = 7 .5 \times 12 \cos (0 .0050 x + 12 t + \frac{π}{4}) \end{aligned}$
$At x = 1 cm and t = 1 \sec :$
$\begin{aligned} v (1, 1) & = 90 \cos (12 .005 + \frac{π}{4}) \\ = 90 \cos (732 {.81}^{\circ}) = 90 \cos (90 \times 8 + 12 {.81}^{\circ}) \\ = 90 \cos (12 {.81}^{\circ}) \\ = 90 \times 0 .975 = 87 .75 cm / s \end{aligned}$
$Now, comparing the given equation with the equation of a propagating wave :$
$y (x, t) = a \sin (kx + ωt + ϕ)$
$So,$
$ω = 12 rad / s$
$k = 0.0050 m^{- 1}$
$∴ v = \frac{12}{0.0050} = 2400 cm / s$
$Hence, the velocity of the wave oscillation at x = 1 cm and t = 1 s$
$is not equal to the velocity of the wave propagation .$
$(b)$
$Propagation constant is given by :$
$k = \frac{2 π}{λ}$
$∴ λ = \frac{2 π}{k} = \frac{2 \times 3 .14}{0 .0050} = 1256 cm = 12 .56 m$
$All the points at distances nλ, i . e . \pm 12.56 m, \pm 25.12 m, . . . where n = \pm 1, \pm 2, . . . and so on$
$will have the same displacement as that of he point at x = 1 cm at t = 2 s, 5 s, and 11 s .$

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