Solution:
Nodes occur where the spatial amplitude term \(sin(2x) = 0\), which implies \(2x = n\pi\) or \(x = \frac{n\pi}{2}\). The closest node to the left end \(x=0\) (where \(n=1\)) is at \(x = \frac{\pi}{2}\text{ cm}\).
A string is fixed at both ends and the vibrations of string is given by the equation \(y = 10sin(2x)cos(2t)\) where \(x, y\) are in cm and \(t\) is in second. Nearby node from left end at \(x = 0\), is at a distance
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