The equation of a transverse wave travelling on a rope is given by $y = 10\sin \pi (0.01x - 2.00t)$ where $y$ and $x$ are in $cm$ and $t$ in $seconds$. The maximum transverse speed of a particle in the rope is about .... $cm/s$
AIIMS 2000, Medium
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(a) The given equation is $y = 10\sin (0.01\pi x - 2\pi t)$
Hence $\omega = $coefficient of $t = 2 \pi$
==> Maximum speed of the particle ${v_{\max }} = a\omega = 10 \times 2\pi $
$= 10 × 2 × 3.14 = 62.8 ≈ 63 cm/s$
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