The maximum acceleration of the body executing SHM is a0 and maximum velocity is v0 the amplitude is given by

Question

The maximum acceleration of the body executing SHM is a0 and maximum velocity is v0 the amplitude is given by (A) (1/a0 v0) (B) (a02/v0) (C) v0 a0 (D) (v02/a0)

Tardigrade Admin · Accepted Answer

For a body executing SHM, maximum acceleration

(a_{0})

is given by

ω^{2} A = a_{0}

...(i)
and maximum velocity

(v_{0})

is

A ω = v_{0}

...(ii)
where

A

is amplitude,

ω

the angular velocity.
Squaring Eq. (ii) and dividing by Eq. (i), we get

A = \frac{v _{0}^{2}}{a _{0}}

Q. The maximum acceleration of the body executing SHM is $a_{0}$ and maximum velocity is $v_{0}$ the amplitude is given by

$\frac{1}{a _{0} v _{0}}$

$\frac{a _{0}^{2}}{v _{0}}$

$v_{0} a_{0}$

$\frac{v _{0}^{2}}{a _{0}}$

For a body executing SHM, maximum acceleration $(a_{0})$ is given by
$ω^{2} A = a_{0}$ ...(i)
and maximum velocity $(v_{0})$ is
$A ω = v_{0}$ ...(ii)
where $A$ is amplitude, $ω$ the angular velocity.
Squaring Eq. (ii) and dividing by Eq. (i), we get
$A = \frac{v _{0}^{2}}{a _{0}}$

Q. The maximum acceleration of the body executing SHM is a0​ and maximum velocity is v0​ the amplitude is given by

a0​v0​1​

v0​a02​​

v0​a0​

a0​v02​​