wave_mechanics critical_thinking Question 3
Question: 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 (\pi x/L)\sin \omega t $ and energy is $ E _{1} $ , and in another experiment its displacement is $ y _{2}=A\sin (2\pi x/L)\sin 2\omega t $ and energy is $ E _{2} $ . Then
[IIT-JEE (Screening) 2001]
Options:
A) $ E _{2}=E _{1} $
B) $ E _{2}=2E _{1} $
C) $ E _{2}=4E _{1} $
D) $ E _{2}=16E _{1} $
Show Answer
Answer:
Correct Answer: C
Solution:
Energy (E) µ (Amplitude)2 (Frequency)2 Amplitude is same in both the cases, but frequency 2$ \omega $ in the second case is two times the frequency $ (\omega ) $ in the first case.
Hence $ E _{2}=4E _{1} $ .
BETA
