# A travelling harmonic wave on a string is described by y(x,t)=7.5sin(0.0050x+12t+Ï€/4) . What are the displacement and velocity of oscillation of a point at x=1 cm and t=1 s? Is this velocity equal to the velocity of wave propagation ?

$y=7.5 \sin (0.0050 x+12 t+\pi / 4)\\ \quad A t \ x=1 \mathrm{cm}\ and\ t=1 s, displacement \\ \begin{array}{l} \text { If } y=7.5 \sin (0.0050 \times 1+12 \times 1+\pi / 4) \\ \text { or } y=7.5 \sin (12.7904 \mathrm{rad}) \\ \begin{array}{l} y=7.5 \sin \left(12.7904 \times \frac{180^{\circ}}{\pi}\right) \\ =7.5 \sin 732.83^{\circ} \\ y=7.5 \sin \left(720^{\circ}+12.83^{\circ}\right) \\ =7.5 \sin 12.83^{\circ} \\ \mathrm{v}_{p}=\frac{\mathrm{dy}}{\mathrm{dt}} \\ \mathrm{v}_{p}=7.5 \times 12 \times \cos (0.0050 \mathrm{x}+12 \mathrm{t}+\pi / 4) \\ =90 \cos (0.0050 \mathrm{x}+12 \mathrm{t}+\pi / 4) \end{array} \end{array}$

$\\ \begin{array}{l} \mathrm{v}_{\mathrm{p}}=\left(90 \cos 12.83^{\circ}\right) \mathrm{cm} / \mathrm{s} \\ =(90 \times 0.9751) \mathrm{cm} / \mathrm{s}=87.75 \mathrm{cm} / \mathrm{s} \end{array} \\ \text{The general equation of the travelling harmonic wave (travelling towards left) is} \\ y=A \sin \left[\omega t+k x+\phi_{0}\right] \\ Comparing\ eqns. (i) and (ii), we get \\ \begin{array}{l} \mathrm{A}=7.5 \mathrm{cm}, \omega=12 \mathrm{rad} / \mathrm{s} \\ \mathrm{k}=0.0050 \mathrm{rad} / \mathrm{cm} \end{array} \\ Speed\ of \ wave\ propagation, \\ \mathrm{v}=\frac{\omega}{\mathrm{k}}=\frac{12 \mathrm{rad} / \mathrm{s}}{0.0050 \mathrm{rad} / \mathrm{cm}} \\ =2400 \mathrm{cm} / \mathrm{s}=-24 \mathrm{m} / \mathrm{s} \\ \text{Negative sign has been taken as the wave travels towards left, We thus find that} v_p\ (particle\ speed) \ is \ not \ equal\ to\ v\ (wave\ speed)$

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