Question

A horizontal platform moves up and down simple harmonically, the total vertical movement being 10 cm. What is the shortest period permissable , if objects resting on the platform are to remain in contact with it throughout the motion? Take g = 980 cm/s^2

Answer 1

Given:

A horizontal platform moves up and down simple harmonically,

total vertical movement = 10 cm

To Find:

Shortest time period permissible.

Solution:

draw the Free body diagram of platform.

A total of 3 forces - 1) normal reaction due to object placed over it (downwards).

2) weight = mg

3)simple harmonic force = mwA(upwards)

where m is the mass of platform,  A is the amplitude of motion and w is the velocity.

A = 5cm (since total vertical movement is 10 cm)

 net force = 0

mwA = mg +N

where N is normal reaction due to object.

The object has to remain in contact with the platform

Hence the threshold value of frequency or time period will be when the object is about to lose contact that is N =0

substitute in equation.

mwA=mg

wA=g

$w= 2\pi f$

f= 1/T

$2\pi fA= g$

substitute  the values to get f

f=1/T

T= .032seconds.

Minimum time period possible = .032 seconds