A string fixed at one end is vibrating in its second overtone. length of string is 10 cm and m

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14.Waves and Sound

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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

A

$\sqrt 3\ mm$

B

$\sqrt 2\ mm$

C

$\frac{{\sqrt 3 }}{2}$

D

None of these

Solution

Second overtone $\Rightarrow 5$ th harmonic

So, $\lambda=\frac{4 L}{5}=8 \mathrm{\,cm}$

$\mathrm{A}_{\mathrm{s}}=(2 \mathrm{\,mm}) \sin (\mathrm{kx})=(2 \mathrm{\,mm}) \sin \left(\frac{2 \pi}{\lambda} x\right)$

$=(2 \mathrm{\,mm}) \sin \left(\frac{2 \pi}{8 \mathrm{\,cm}} \times 1 \mathrm{\,cm}\right)$

$=(2 \mathrm{\,mm}) \sin \left(\frac{\pi}{4}\right)=\sqrt{2} \mathrm{\,mm}$

Standard 11

Physics

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