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  • The two ends of an stretched wire of length L are fixed at x = 0 and x = L. In one experiment the displacement is <img alt="y_1=A \sin \left(\frac{\pi x}{L}\right) \sin \omega t" src="https://

The two ends of an stretched wire of length L are fixed at x = 0 and x = L. In one experiment the displacement is y_1=A \sin \left(\frac{\pi x}{L}\right) \sin \omega t and energy is E1 and in another experiment its displacement y_2=A \sin \left(\frac{2\pi x}{L}\right) \sin 2 \omega t is and energy is E2 then 

Option: 1

E2 = E1  


Option: 2

E2 = 2 E1  


Option: 3

E2 = 4 E1  


Option: 4

E2 = 16 E1  


Answers (1)

best_answer

E = energy density × volume

E = \frac{1}{2} \rho \omega^2 A^2 S L

or E\, \alpha\, \omega^2, rest all quantities are common for both the waves.

\omega_2=2 \omega \text { and } \omega_1=\omega

E2 = 4 E1

 

 

Posted by

Gaurav

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