Wave Mechanics Question 534
Question: A particle performs SHM in a straight line. In the first second, starting from rest, it travels a distance a and in the next second it travels a distance b in the same direction. The amplitude of the SHM is
Options:
A) $ a-b $
B) $ \frac{2a-b}{3} $
C) $ \frac{2a^{2}}{3a-b} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] When particle starts from mean position, $ x=A,\cos ,\omega t $
$ \
Therefore ,(A-a)=A,\cos (\omega \times 1) $ Or $ (A-a)=A,cos,\omega $ ?..(i) $ Also,A-(a+b)=A,\cos (\omega \times 2) $ Or $ A-(a+b)=A,\cos ,2\omega $ ??(ii) After simplifying above equations, we get $ A=[ \frac{2a^{2}}{3a-b} ] $
BETA
