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. In Moseley’s law √ν=a(z-b), the values of the screening constant for K-series
  2. A particle starts its motion from rest under the action of a constant force
  3. The temperature of a metal block is increased from 27⁰C to 84⁰C. The rate of radiated energy
  4. The dimensional formula of Plancks’s constant and angular momentum are
  5. A small rod of bismuth is suspended freely between the poles of a strong electromagnet.