A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the…

1. The number of nodes is 5.
2. The length of the string is 0.25 m.
3. The maximum displacement of the midpoint of the string, from its equilibrium position is 0.01 m.
4. The fundamental frequency is 100 Hz.

Solution

​
$\text{Nodes}=\mathrm{6\; }$
$\lambda =\text{0.1}$
$\therefore 5^{\text{th}}harmonic, 5\frac{\lambda }{2}=\text{L}$
$\text{L}=\text{0.25}$
$\text{At}\frac{\text{L}}{2}$
${\text{y}}_{\text{max}}=\text{0.01 sin in }\left(\text{62.8}\times \frac{\text{0.25}}{2}\right)$
          $=\text{0.01}$
$\text{Interfering wave velocity}=\frac{\text{w}}{\text{k}}=\frac{628}{\text{62.8}}=10$
$\therefore \text{fundamental frequency}=\frac{\nu }{2\text{L}}=20$ `

Asked in: JEE Advanced 2013 (Paper 1)

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