A small block is placed on a wooden plank which is undergoing SHM in vertical direction with a period of T.What is the maximum amplitude of motion so the block does not leave the plank?

Options

(a) #(g*T^2)/pi^2#
(b) #(g*T^2)/(4pi^2)#
(c)#(4*g*T^2)/pi^2#
(d) none of these

1 Answer
Mar 26, 2018

I get (b)

Explanation:

SHM is described by the expression

#y=Asinomegat#
where #A# is maximum amplitude and #omega=(2pi)/T# is the angular frequency.

We also have

Velocity #v=doty=omegaAcosomegat#
Acceleration #a= dotv=-omega^2Asinomegat=-omega^2y#

The block will not leave the plank so long as its gravitational force is less than centrifugal force experienced by it due to circular motion. As such we have the condition

#m|vecg|=m|veca|#
where #m# is mass of the block.
#=>|vecg|=|veca|#
#=>g=omega^2y#

For maximum acceleration #y=A#. Therefore we get

#g=omega^2A#
#=>A=g/(omega^2)#
#=>A=g/(((2pi)/T)^2)#
#=>A=(gT^2)/(4pi^2)#