The equation of a transverse wave travelling along a string is $y(x, t)=4.0 \sin \left[20 \times 10^{-3} x+600 t\right] \mathrm{mm}$, where $x$ is in mm and $t$ is in second. The velocity of the wave is :

<p>Let's analyze the wave equation step by step.</p> <p><p>The given wave is:</p> <p>$y(x, t) = 4.0 \sin \left[20 \times 10^{-3} x + 600 t\right] \text{ mm}.$</p></p> <p><p>First, simplify the coefficient of $ x $:</p> <p>$20 \times 10^{-3} = 0.02 \, \text{mm}^{-1}.$</p> <p>So the equation becomes:</p> <p>$y(x, t) = 4.0 \sin\left(0.02x + 600t\right) \text{ mm}.$</p></p> <p><p>A standard form for a travelling wave is:</p> <p>$y(x, t) = A \sin(kx - \omega t)$</p> <p>which represents a wave moving in the positive $ x $-direction with speed $ v = \frac{\omega}{k} $.</p></p> <p><p>Notice that our wave equation has the form:</p> <p>$\sin(0.02x + 600t)$</p> <p>The positive sign in front of $ 600t $ means we can rewrite the phase as:</p> <p>$0.02x + 600t = 0.02x - (-600t),$</p> <p>which indicates that the angular frequency $ \omega $ in the standard form is effectively $ -600 $.</p></p> <p><p>The velocity $ v $ of a wave is determined from the phase (for a constant phase, $ \phi = $ constant):</p> <p>$k x + \omega t = \text{constant}.$</p> <p>Differentiating with respect to $ t $:</p> <p>$k \frac{dx}{dt} + \omega = 0,$</p> <p>which gives:</p> <p>$\frac{dx}{dt} = -\frac{\omega}{k}.$</p></p> <p><p>Substituting the values:</p></p> <p><p>$ k = 0.02 \, \text{mm}^{-1} $</p></p> <p><p>$ \omega = 600 \, \text{s}^{-1} $</p> <p>We have:</p> <p>$v = -\frac{600}{0.02} = -30000 \text{ mm/s}.$</p></p> <p><p>Convert the velocity from mm/s to m/s:</p> <p>$-30000 \, \text{mm/s} = -30 \, \text{m/s}.$</p></p> <p>Thus, the velocity of the wave is $-30 \, \text{m/s}$.</p> <p>The correct answer is Option D.</p>