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

Question

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 (A) 63 cm/s (B) 75 cm/s (C) 100 cm/s (D) 121 cm/s

Tardigrade Admin · Accepted Answer

The given equation is y=10 sin π(0.01 π x-2 π t) Hence ω= coefficient of t=2 π ⇒ Maximum speed of the particle v max =a ω=10 × 2 π =10 × 2 × 3.14=62.8 ≈ 63 cm / s

Q. 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 $c m$ and $t$ in seconds. The maximum transverse speed of a particle in the rope is about

63 cm/s

75 cm/s

100 cm/s

121 cm/s

The given equation is $y = 10 sin π (0.01 π x - 2 π t)$
Hence $ω =$ coefficient of $t = 2 π$
$\Rightarrow$ Maximum speed of the particle
$v_{m a x} = aω = 10 \times 2 π$
$= 10 \times 2 \times 3.14 = 62.8 \approx 63 c m / s$

Q. The equation of a transverse wave travelling on a rope is given by y=10sinπ(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