Equation of a transverse wave travelling in a rope is given by $y = 5 \sin(4.0 t - 0.02 x)$, where $y$ and…
Equation of a transverse wave travelling in a rope is given by $y = 5 \sin(4.0 t - 0.02 x)$, where $y$ and $x$ are expressed in cm and time in seconds. Calculate (i) the amplitude, frequency, velocity and wavelength of the wave. (ii) the maximum transverse velocity speed and acceleration of a particle in the rope.
Solution
Sol. (i) Comparing the given equation with the standard equation of wave motion, $y = A\sin\left(2\pi ft - \dfrac{2\pi}{\lambda}x\right)$ where, A, f and $\lambda$ are amplitude, frequency and wavelength, respectively. Thus, amplitude, $A = 5\ \text{cm}$, $2\pi f = 4$
$\Rightarrow$ Frequency, $f = \dfrac{4}{2\pi} = 0.637\ \text{Hz}$
Again, $\dfrac{2\pi}{\lambda} = 0.02$
$\Rightarrow$ Wavelength, $\lambda = \dfrac{2\pi}{0.02} = 100\pi\ \text{cm}$
Velocity of the wave, $v = f\lambda = \dfrac{4}{2\pi}\cdot\dfrac{2\pi}{0.02} = 200\ \text{cm s}^{-1}$
(ii) Transverse velocity of the particle,
$u = \dfrac{dy}{dt} = 5 \times 4 \cos\left(4.0\ t - 0.02\ x\right)$
$= 20\cos\left(4.0\ t - 0.02x\right)$
Maximum velocity of the particle = $20\ \text{cm s}^{-1}$
Particle acceleration,
$a = \dfrac{d^2y}{dt^2} = -20 \times 4 \sin\left(4.0\ t - 0.02\ x\right)$
Maximum particle acceleration = $80\ \text{cm s}^{-2}$.
Answer: Frequency $f=0.637\ \text{Hz}$; Wavelength $\lambda=100\pi\ \text{cm}$; Wave velocity $v=200\ \text{cm s}^{-1}$; Maximum particle velocity $=20\ \text{cm s}^{-1}$; Maximum particle acceleration $=80\ \text{cm s}^{-2}$