A flat horizontal board moves up and down under S.H.M. vertically with amplitude A. The shortest permissible time period of the vibration such that an object placed on the board may not lose contact with the board is

Question

A flat horizontal board moves up and down under S.H.M. vertically with amplitude A. The shortest permissible time period of the vibration such that an object placed on the board may not lose contact with the board is (A) 2 π √(g/A) (B) 2 π √(A/g) (C) 2 π √(2 A/g) (D) (π/2) √(A/g)

Tardigrade Admin · Accepted Answer

Maximum acceleration of the system (a max )=-ω2 A For a block to escape the board the acceleration must be equal to 9 at the top-most point. g=ω2 A ω=√(g/A) Time period =(2 π/ω)=√(A/g)

Q. A flat horizontal board moves up and down under S.H.M. vertically with amplitude $A$ . The shortest permissible time period of the vibration such that an object placed on the board may not lose contact with the board is

$2 π \frac{g}{A}$

$2 π \frac{A}{g}$

$2 π \frac{2 A}{g}$

$\frac{π}{2} \frac{A}{g}$

Maximum acceleration of the system $(a_{m a x}) = - ω^{2} A$
For a block to escape the board the acceleration must be equal to $9$ at the top-most point.
$g = ω^{2} A$
$ω = \frac{g}{A}$
Time period $= \frac{2 π}{ω} = \frac{A}{g}$

Q. A flat horizontal board moves up and down under S.H.M. vertically with amplitude A. The shortest permissible time period of the vibration such that an object placed on the board may not lose contact with the board is

2πAg​​

2πgA​​

2πg2A​​

2π​gA​​