1. Tardigrade
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  3. Physics
  4. The equation of a transverse wave travelling on a rope is given by y=10 sin π(0.01 x-2.00 t) where y and x are in cm and t in seconds. The maximum transverse speed of a particle in the rope is about

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Q. The equation of a transverse wave travelling on a rope is given by $y=10\, \sin \pi(0.01 x-2.00 t)$ where $y$ and $x$ are in $cm$ and $t$ in seconds. The maximum transverse speed of a particle in the rope is about

Waves

Solution:

The given equation is $y=10\, \sin \pi(0.01\, \pi x-2 \pi t)$
Hence $\omega=$ coefficient of $t=2\, \pi$
$\Rightarrow $ Maximum speed of the particle
$v_{\max }=a \omega=10 \times 2 \pi$
$=10 \times 2 \times 3.14=62.8 \approx 63\, cm / s$