A particle executes linear simple harmonic motion with an amplitude of 2 cm. When the particle is at 1 cm from the mean position the magnitude of its velocity is equal to that of its acceleration. Then its time period in second is

Question

A particle executes linear simple harmonic motion with an amplitude of 2 cm. When the particle is at 1 cm from the mean position the magnitude of its velocity is equal to that of its acceleration. Then its time period in second is (A)  (1/2π √3)  (B)  2π √3  (C)  (2π /√3)  (D)  (√3/2π )

Tardigrade Admin · Accepted Answer

Velocity = acceleration (Given) ∴ ω √a2-y2=ω2 y √(2)2-(1)2=ω(1) ⇒ ω=√3 T=(2 π/ω) ⇒ T=(2 π/√3)

Q. A particle executes linear simple harmonic motion with an amplitude of $2 c m$ . When the particle is at $1 c m$ from the mean position the magnitude of its velocity is equal to that of its acceleration. Then its time period in second is

$\frac{1}{2 π 3}$

$2 π 3$

$\frac{2 π}{3}$

$\frac{3}{2 π}$

Velocity $=$ acceleration (Given)
$∴ ω a^{2} - y^{2} = ω^{2} y$
$(2)^{2} - (1)^{2} = ω (1)$
$\Rightarrow ω = 3$
$T = \frac{2 π}{ω}$
$\Rightarrow T = \frac{2 π}{3}$

Q. A particle executes linear simple harmonic motion with an amplitude of 2cm. When the particle is at 1cm from the mean position the magnitude of its velocity is equal to that of its acceleration. Then its time period in second is

2π3​1​

2π3​

3​2π​

2π3​​