The vibrations of a string fixed at both ends are described by the equation $y = (5.00\ \text{mm})\sin[(1…
The vibrations of a string fixed at both ends are described by the equation
$y = (5.00\ \text{mm})\sin[(1.57\ \text{cm}^{-1})x]\sin[(314\ \text{s}^{-1})t]$
If the length of the string is 10.0cm, locate the nodes and the antinodes. How many loops are formed in the vibration?
Solution
Sol. The given equation can be written as
$y = 5\sin(1.57x)\sin(314t)$
$\quad = 5\sin\left(\dfrac{\pi x}{2}\right)\sin(100\pi t)\qquad\ldots(i)$
General equation of standing wave,
$y = 2A\sin(kx)\sin(\omega t)\qquad\ldots(ii)$
Comparing both the equations, we get
$k = \dfrac{\pi}{2}\;\text{cm}^{-1},\; \omega = 100\pi\;\text{rad s}^{-1}$
$\therefore\; k = \dfrac{2\pi}{\lambda} = \dfrac{\pi}{2} \Rightarrow \lambda = 4\text{ cm}$
$v = \dfrac{\omega}{k} = \dfrac{100\pi}{\pi/2} = 200\text{ cm s}^{-1} = 2\text{ m s}^{-1}$
$\therefore\;$ Length, $L = n\dfrac{\lambda}{2} \Rightarrow 10 = n\times \dfrac{4}{2} \Rightarrow n = 5$
Hence, string is vibrating in 5 loops.
Answer: $5\ \text{loops}$