The equation given below represents a stationary wave set up in a medium $y = 12\sin(4\pi x)\sin(40\pi t)$…
The equation given below represents a stationary wave set up in a medium
$y = 12\sin(4\pi x)\sin(40\pi t)$
where, y and x are in cm and t is in second. Calculate the amplitude, wavelength and velocity of the component waves
Solution
Sol. On compare the given equation with stationary wave equation, $y = 2A\sin kx\sin\omega t$, we get
$2A = 12\text{cm},\; k = 4\pi\;\text{cm}^{-1}$
and
$\omega = 40\pi\;\text{rad s}^{-1}$
Therefore, amplitude of the component wave = $6\text{ cm}$
As,
$k = 4\pi \Rightarrow \dfrac{2\pi}{\lambda} = 4\pi$
$\Rightarrow \lambda = \dfrac{1}{2} = 0.5\text{cm}$
Further, velocity of the wave,
$v = \dfrac{\omega}{k} = \dfrac{40\pi}{4\pi} = 10\text{ cm s}^{-1}$
Answer: $10\ \text{cm s}^{-1}$