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 : 

 

  • Option 1)

    A _0 + \sqrt{ A^2 + B^2 }

  • Option 2)

    \sqrt{ A^2 + B^2 }

  • Option 3)

    \sqrt{A_0 ^2+ (A + B)^2 }

  • Option 4)

    A+ B

Answers (1)

The particle execute SHM about its center at \alpha = A_0 \\\\ amplitude = \sqrt { A^2 + B^2 }


Option 1)

A _0 + \sqrt{ A^2 + B^2 }

Option 2)

\sqrt{ A^2 + B^2 }

Option 3)

\sqrt{A_0 ^2+ (A + B)^2 }

Option 4)

A+ B

Most Viewed Questions

Preparation Products

Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports,.

₹ 29999/- ₹ 24999/-
Buy Now
Knockout NEET 2025

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 45000/-
Buy Now
Knockout NEET (One Month)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports.

₹ 7999/- ₹ 4999/-
Buy Now
Knockout NEET 2023 (One Month)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports.

₹ 13999/- ₹ 6999/-
Buy Now
Knockout NEET (Three Months Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 5999/- ₹ 4499/-
Buy Now
Boost your Preparation for NEET 2021 with Personlized Coaching
 
Exams
Articles
Questions