Wave Motion 5 Question 4
4. The ends of a stretched wire of length $L$ are fixed at $x=0$ and $x=L$. In one experiment the displacement of the wire is $y _1=A \sin \left(\frac{\pi x}{L}\right) \sin \omega t$ and energy is $E _1$ and in other experiment its displacement is $y _2=A \sin \left(\frac{2 \pi x}{L}\right) \sin 2 \omega t$ and energy is $E _2$. Then
(2001, 1M)
(a) $E _2=E _1$
(b) $E _2=2 E _1$
(c) $E _2=4 E _1$
(d) $E _2=16 E _1$
Show Answer
Answer:
Correct Answer: 4. (c)
Solution:
- Energy $E \propto$ (amplitude) $^{2}$ (frequency $)^{2}$
Amplitude $(A)$ is same in both the cases, but frequency $2 \omega$ in the second case is two times the frequency $(\omega)$ in the first case.
Therefore,
$$ E _2=4 E _1 $$
