A string fixed at one end is vibrating in its second overtone. The length of the string is $10\ cm$ and maximum amplitude of vibration of particles of the string is $2\ mm$ . Then the amplitude of the particle at $9\ cm$ from the open end is
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Second overtone $\Rightarrow 5$ th harmonic
So, $\lambda=\frac{4 L}{5}=8 \mathrm{\,cm}$
$A_{s}=(2\, m m) \sin (k x)=(2\, m m) \sin \left(\frac{2 \pi}{\lambda} x\right)$
${=(2 \mathrm{\,mm}) \sin \left(\frac{2 \pi}{8 \mathrm{cm}} \times 1 \mathrm{cm}\right)} $
${=(2 \mathrm{\,mm}) \sin \left(\frac{\pi}{4}\right)=\sqrt{2} \mathrm{\,mm}}$
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