Question

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 $\text{seconds}$. The maximum transverse speed of a particle $\left(\text{in}cm/s\right)$ in the rope is about (Round off the answer to nearest integer)

Answer 1

The given equation is
$y=10\sin (0.01\pi x-2\pi t)$
The standard equation can be written as
$y=A\sin (kx-\omega t)$
Hence $\omega =\text{coefficient of}t=2\pi$
$\Rightarrow$ Maximum speed of the particle
$v_{max}=a\omega =10\times 2\pi =10\times 2\times 3.14=62.8$$\approx 63cm/s$
Final answer: $\left(63\right)$