JEE MainPhysicsOscillations
A particle executes simple harmonic motion. The area enclosed by its velocity-displacement graph is S , and the maximum velocity of the particle is v₀ . The maximum acceleration of the particle is
Options
- Av₀^3 S
- Bv₀^2 S
- Cv₀^3 S
- DS v₀
Correct answer
C. v₀^3 S
Step-by-step solution
The relationship between velocity v and displacement x for a particle in simple harmonic motion is: x^2 A^2 + v^2 (A )^2 = 1 This is the equation of an ellipse with semi-axes A and A . The area S of this ellipse is given by: S = (A)(A ) Since the maximum velocity is v₀ = A , we can write the area as: S = A v₀ From this, the amplitude A is: A = S v₀ The maximum acceleration a₀ of the particle is given by a₀ = ^2 A . We can rewrite this in terms of v₀ and A : a₀ = (A )^2 A = v₀^2 A Substituting the expression for A :