Answer :
Answer:
maximum amplitude = 0.13 m
Explanation:
Given that
Time period T= 0.74 s
acceleration of gravity g= 10 m/s²
We know that time period of simple harmonic motion given as
[tex]T=\dfrac{2\pi}{\omega}[/tex]
[tex]0.74=\dfrac{2\pi}{\omega}[/tex]
ω = 8.48 rad/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
10= 8.48² A
A=0.13 m
maximum amplitude = 0.13 m
The maximum amplitude of the motion for which the block does not separate from the plate is : 0.14 m
Given data :
period ( T ) = 0.74 secs
acceleration due to gravity ( g ) = 9.8 m/s62
Determine the maximum amplitude of the motion
Time period of simple harmonic motion ( T ) = [tex]\frac{2\pi }{w}[/tex]
First step : solve for w
w = 2π / T
= 2π / 0.74
= 8.49 rad/sec
Next step : determine the maximum amplitude ( A )
a = w² A
where ; a = maximum acceleration = 9.8 , w = 8.49
therefore
A = a / w²
= 9.8 / ( 8.49 )²
= 0.136 ≈ 0.14 m
Hence we can conclude that The maximum amplitude of the motion for which the block does not separate from the plate is : 0.14 m.
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