A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is :

Question

A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is : (A) (√5/2 π ) (B) (4 π /√5) (C) (2 π /√3) (D) (√5/ π )

Tardigrade Admin · Accepted Answer

v=ω √A2-x2 a=x ω2 v=a ω √A2-x2=x ω2 √(3)2-(2)2=2((2 π/T)) √5=(4 π/T) T=(4 π/√5)

Q. A particle executes linear simple harmonic motion with an amplitude of $3 c m$ . When the particle is at $2 c m$ from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is :

$\frac{5}{2 π}$

$\frac{4 π}{5}$

$\frac{2 π}{3}$

$\frac{5}{π}$

$v = ω A^{2} - x^{2}$
$a = x ω^{2}$
$v = a$
$ω A^{2} - x^{2} = x ω^{2}$
$(3)^{2} - (2)^{2} = 2 (\frac{2 π}{T})$
$5 = \frac{4 π}{T}$
$T = \frac{4 π}{5}$

Q. A particle executes linear simple harmonic motion with an amplitude of 3cm. When the particle is at 2cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is :

2π5​​

5​4π​

3​2π​

π5​​