Wave motionHardBloom L3

Question

A string of length 3L is fixed at both ends. It resonates with a tuning fork in third harmonic with amplitude at antinode equal to A0. At time t = 0, a string element at position of antinode is at half its positive amplitude and moving towards mean position. Displacement of a string element at L/2 is given by

Options

A.$\frac{A_{0}}{2}\sin\left( \omega t + \frac{11\pi}{6} \right)$
B.$\frac{\sqrt{3}A_{0}}{2}\sin\left( \omega t + \frac{5\pi}{6} \right)$
C.$A_{0}\sin\left( \omega t + \frac{5\pi}{6} \right)$
D.$\frac{A_{0}}{2}\sin\left( \omega t + \frac{5\pi}{6} \right)$

Solution

Sol. Third harmonic ⇒ 3 loops

⇒ 3L = $\frac{3\lambda}{2}$

⇒ K = $\frac{2\pi}{\lambda} = \frac{\pi}{L}$

Let equation be

y = A sin kx sin (ωt + φ)

At t = 0, y = $\frac{A_{0}}{2}$ at antinode & moving towards mean

⇒φ = π – $\frac{\pi}{6} = \frac{5\pi}{6}$

So, at $\frac{L}{2}$

y = A0 sin$\left( \omega t + \frac{5\pi}{6} \right)$

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