In S.H.M. if the magnitude of maximum acceleration is a and maximum velocity is β, then its time period will be

Question

In S.H.M. if the magnitude of maximum acceleration is a and maximum velocity is β, then its time period will be (A)  (2 π α/β) (B)  (2 π β/α) (C)  (α/2 π β) (D)  ( β/2 πα)

Tardigrade Admin · Accepted Answer

Maximum acceleration, α = ω2 A Maximum velocity, β = ω A ∴ : :(α/β) =(ω2 A/ω A) =ω = (2π/T) or ,T = (2πβ/α)

Q. In S.H.M. if the magnitude of maximum acceleration is a and maximum velocity is $β$ , then its time period will be

$\frac{2 π α}{β}$

$\frac{2 π β}{α}$

$\frac{α}{2 π β}$

$\frac{β}{2 π α}$

Maximum acceleration, $α = ω^{2} A$
Maximum velocity, $β = ω A$
$∴ \frac{α}{β} = \frac{ω ^{2} A}{ω A} = ω = \frac{2 π}{T}$
or , $T = \frac{2 π β}{α}$

Q. In S.H.M. if the magnitude of maximum acceleration is a and maximum velocity is β, then its time period will be

β2πα​

α2πβ​

2πβα​

2παβ​