A particle is executing a simple harmonic motion. Its maximum acceleration

A particle is executing a simple harmonic motion. Its maximum acceleration is α and maximum velocity is β. Then, its time period of vibration will be

Options

(a) β²/α
(b) 2πβ/α
(c) β²/α²
(d) α/β

Correct Answer:

2πβ/α

Explanation:

As, we know, in Simple Harmonic Motion
Maximum acceleration of the particle, α = Aω²
Maximum velocity, β = Aω
⇒ ω = α / β
⇒ T = 2π / ω = 2πβ / α [Since, ω = 2π / T].

admin:

Related Questions

  1. An electron in a circular orbit of radius 0.05 nm performs 10¹⁶ rev/s. The magnetic
  2. A body moves with a uniform acceleration a and zero initial velocity
  3. For a transformer, the turns ratio is 3 and its efficiency is 0.75. The current flowing
  4. In a coil having e.m.f. of 8 V induced for 0.05 s. The value of current changes
  5. A body is thrown vertically upward in air when air resistance is taken into account