A wave disturbance in a medium is described by $y(x,\,t) = 0.02\cos \,\left( {50\,\pi t + \frac{\pi }{2}} \right)\cos (10\pi x)$, where $ x$ and $y$ are in metres and $t$ in seconds
IIT 1995, Diffcult
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(d) $y = 0.02\cos (10\,\pi x)\cos \,\left( {50\,\pi \,t + \frac{\pi }{2}} \right)$
At node, amplitude $= 0$
$ \Rightarrow $ $\cos (10\pi x) = 0 \Rightarrow 10\,\pi x = \frac{\pi }{2},\frac{{3\pi }}{2}$
$ \Rightarrow \,x = \frac{1}{{20}} = 0.05 m, 0.15m …..$
At antinode, amplitude is maximum
$ \Rightarrow $$\cos (10\pi x) = \pm 1 \Rightarrow x = 0,\pi ,2\pi ...$
$==> x = 0, 0.1m, 0.2m …$
Now $\lambda = 2 \times $ Distance between two nodes or antinodes
= $2 \times 0.1 = 0.2 m$ and $\frac{{2\pi vt}}{\lambda } = 50\pi t$
$v = 25\lambda = 25 \times 0.2 = 5m/\sec $.
At node, amplitude $= 0$
$ \Rightarrow $ $\cos (10\pi x) = 0 \Rightarrow 10\,\pi x = \frac{\pi }{2},\frac{{3\pi }}{2}$
$ \Rightarrow \,x = \frac{1}{{20}} = 0.05 m, 0.15m …..$
At antinode, amplitude is maximum
$ \Rightarrow $$\cos (10\pi x) = \pm 1 \Rightarrow x = 0,\pi ,2\pi ...$
$==> x = 0, 0.1m, 0.2m …$
Now $\lambda = 2 \times $ Distance between two nodes or antinodes
= $2 \times 0.1 = 0.2 m$ and $\frac{{2\pi vt}}{\lambda } = 50\pi t$
$v = 25\lambda = 25 \times 0.2 = 5m/\sec $.
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