Question

A uniform string of length $l$, fixed at both ends is vibrating in its $2nd$ overtone. The maximum amplitude is '$a$' and tension in string is '$T$', if the energy of vibration contained between two consecutive nodes is $\frac{K}{8}.\frac{a^2{\pi }^{2}T}{l}$ then '$K$' is

Answer 1

Total Energy $E={\int }_o^t\frac{1}{2}dm{v}^2$$={\int }_o^t\frac{1}{2}dm.A_x^2{\omega }^2={\int }_0^t\frac{1}{2}\left(\frac{m}{l}\right)dx.a^2{\sin }^{2}kx.{\omega }^2=\frac{1}{4}m{a}^2{\omega }^2$

$\omega =2\pi f=2\pi .\frac{3v}{2l}\sqrt{\frac{T}{\mu }}=\frac{3\pi }{l}\sqrt{\frac{Tl}{m}}$

$\therefore Energy=\frac{1}{4}m{a}^2.\frac{9{\pi }^2}{l^2}.\frac{Tl}{m}$

Energy $=\frac{9}{4}\frac{a^2{\pi }^{2}T}{l}$

So, energy between two consecutive nodes $=\frac{3}{4}\frac{a^2{\pi }^{2}T}{l}$

$=>K=6$