NEET Solved Paper 2019 Question 11
Question: The displacement of a particle executing simple harmonic motion is given by $ y=A _0+A,sin\omega t+B,cos\omega t $ . Then the amplitude of its oscillation is given by: [NEET 5-5-2019]
Options:
A) $ \sqrt{A_0^{2}+{{(A+B)}^{2}}} $
B) A+B
C) $ A _0=\sqrt{A^{2}+B^{2}} $
D) $ \sqrt{A^{2}+B^{2}} $
Show Answer
Answer:
Correct Answer: D
Solution:
- $ y=A _0+Asin\omega t+Bcos\omega t $ Amplitude of $ Asin\omega t+Bcos\omega t=\sqrt{A^{2}+B^{2}} $
