A particle moves so that its acceleration a is given by a=-bx , where x is displacement from equilibrium position and b is a non-negative real constant. The time period of oscillation of the particle is

Question

A particle moves so that its acceleration a is given by a=-bx , where x is displacement from equilibrium position and b is a non-negative real constant. The time period of oscillation of the particle is (A) 2π √b (B) (2 π /b) (C) (2 π /√b) (D) 2√(π /b)

Tardigrade Admin · Accepted Answer

The relation between acceleration ( a ) and displacement ( x ) for a body in SHM is a=-ω 2x Given, a=-bx On comparing the two equations, we get ω 2=b ∴ ω =√b Since, ω =(2 π /T) ∴ (2 π /T)=√b ⇒ T=(2 π /√b)

Q. A particle moves so that its acceleration $a$ is given by $a = - b x$ , where $x$ is displacement from equilibrium position and $b$ is a non-negative real constant. The time period of oscillation of the particle is

$2 π b$

$\frac{2 π}{b}$

$\frac{2 π}{b}$

$2 \frac{π}{b}$

The relation between acceleration ( $a$ ) and displacement ( $x$ ) for a body in SHM is
$a = - ω^{2} x$
Given, $a = - b x$
On comparing the two equations, we get
$ω^{2} = b$
$∴$ $ω = b$
Since, $ω = \frac{2 π}{T}$
$∴$ $\frac{2 π}{T} = b$
$\Rightarrow T = \frac{2 π}{b}$

Q. A particle moves so that its acceleration a is given by a=−bx , where x is displacement from equilibrium position and b is a non-negative real constant. The time period of oscillation of the particle is

2πb​

b2π​

b​2π​

2bπ​​

The relation between acceleration ( a ) and displacement ( x ) for a body in SHM is a=−ω2x Given, a=−bx On comparing the two equations, we get ω2=b ∴ ω=b​ Since, ω=T2π​ ∴ T2π​=b​ ⇒T=b​2π​

Q. A particle moves so that its acceleration $a$ is given by $a = - b x$ , where $x$ is displacement from equilibrium position and $b$ is a non-negative real constant. The time period of oscillation of the particle is

$2 π b$

$\frac{2 π}{b}$

$\frac{2 π}{b}$

$2 \frac{π}{b}$

The relation between acceleration ( $a$ ) and displacement ( $x$ ) for a body in SHM is
$a = - ω^{2} x$
Given, $a = - b x$
On comparing the two equations, we get
$ω^{2} = b$
$∴$ $ω = b$
Since, $ω = \frac{2 π}{T}$
$∴$ $\frac{2 π}{T} = b$
$\Rightarrow T = \frac{2 π}{b}$