Question

A block rests on a ?at plate that executes vertical simple harmonic motion with a period of 0.58s.

What is the maximum amplitude of the motion for which the block does not separate from the plate?

The acceleration of gravity is 9.8 m/s^2.

Answer 1

Given Information:

Period = T = 0.58 seconds

Acceleration due to gravity = g = 9.8 m/s²

Required Information:

Maximum amplitude = A = ?

Answer:

Maximum amplitude = 0.083 m

Step-by-step explanation:

The relation between the period and angular velocity of a simple harmonic motion is given by

ω = 2π/T

Whereas the maximum acceleration is given by

a = ω²A

Where A is the corresponding maximum amplitude

a = (2π/T)²A

Now will we equate the maximum acceleration with acceleration due to gravity

g = (2π/T)²A

A = g/(2π/T)²

Finally, substitute the given values

A = 9.8/(2π/0.58)²

A = 9.8/117.35

A = 0.083 m

Therefore, the maximum amplitude for which the block doesn't separate from the plate is 0.083 m any increase beyond this value will result in separation.

Answer 2

maximum amplitude = 0.08 m

Step-by-step explanation:

Given that

Time period T= 0.58 s

acceleration of gravity g= 9.8 m/s²

We know that time period of simple harmonic motion given as

T = 2π/ω

0.58 = 2π/ω

ω = 10.83rad/s

ω=angular frequency

Lets take amplitude = A

The maximum acceleration given as

a= ω² A

The maximum acceleration should be equal to g ,then block does not separate

a= ω² A

9.8 = 10.83² A

A = 0.08m

maximum amplitude = 0.08 m