The equation of a transverse wave travelling along a string is y(x, t)=4.0 sin [20 × 10 -3 x+600 t ] m m , where x is in the mm and t is in second. The velocity of the wave is :

Physics · Transverse waves

JEE Main 2025 — 23 January, Evening Shift — Question 51

The equation of a transverse wave travelling along a string is

$y(x, t)=4.0 \sin \left[20 \times 10^{-3} x+600 t\right] m m$ , where x is in the mm

and t is in second. The velocity of the wave is :

$y=4 \sin \left(20 \times 10^{-3} x+600 t\right)$

Here

$\omega =600\text{ }\!\!~\!\!\text{ }{{\text{s}}^{-1}}$

$\text{k}=20\times {{10}^{-3}}\text{ }\!\!~\!\!\text{ m}/{{\text{s}}^{-1}}$

$\therefore \quad \mathrm{v}=\frac{\mathrm{w}}{\mathrm{k}}=\frac{600}{20 \times 10^{-3}}$

$=30 \times 10^{-3} \mathrm{~mm} / \mathrm{s}$

$=30 \mathrm{~m} / \mathrm{s}$

& direction is towards -ve x axis

$\therefore \mathrm{v}=-30 \mathrm{~m} / \mathrm{s}$

Answer key and solution verified before publishing.

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