The displacement of a standing wave on a string is given by $y(x,t)=0.4\sin(0.5x)\cos(30t)$ where, x and y…
The displacement of a standing wave on a string is given by
$y(x,t)=0.4\sin(0.5x)\cos(30t)$
where, x and y are in centimetres.
(i) Find the frequency, amplitude and wave speed of the component waves.
(ii) What is the particle velocity at $x = 2.4\ \text{cm}$ and $t = 0.8\ \text{s}$ ?
Solution
Sol. (i) The given wave can be written as the sum of two component waves as
$y(x,t)=0.2\sin(0.5x-30t)+0.2\sin(0.5x+30t)$
The two component waves are
$y_1(x,t)=0.2\sin(0.5x-30t)$
(Travelling in positive x-direction)
and $y_2(x,t)=0.2\sin(0.5x+30t)$
(Travelling in negative x-direction)
Now, $\omega = 30\ \mathrm{rad\ s}^{-1}$ and $k = 0.5\ \mathrm{cm}^{-1}$
$\therefore$ Frequency, $f = \omega/2\pi = 15/\pi\ \mathrm{Hz}$
Amplitude, $A = 0.2\ \mathrm{cm}$
and wave speed, $v = \omega/k = 30/0.5 = 60\ \mathrm{cm\ s}^{-1}$
(ii) Particle velocity,
$v_P(x,t)=\dfrac{dy}{dt} = -12\sin(0.5x)\sin(30t)$
$\therefore\ v_P\ (x = 2.4\ \mathrm{cm},\ t = 0.8\ \mathrm{s})$
$= -12\sin(1.2)\sin(24)$
$= 10.12\ \mathrm{cm\ s}^{-1}$
Answer: $10.12\ \mathrm{cm\ s}^{-1}$