Answer:
The amplitude of oscillation is <u>0.07698m</u>
Explanation:
The given motion is an example of Simple Harmonic Motion (SHM).
For a simple harmonic spring-block motion the angular frequency (w) is given by -
w =
-- 1
,where k=spring constant of the spring and m=mass of the block.
We know that time period of SHM is given by -
T =
= 0.467
∴ w =
=
= 13.45 
The general equation of motion -
x = A sin(wt+∅) , where A : Amplitude of oscillation
t : time
∅ : phase difference of oscillation
x : Displacement of block
∴ v = Aw cos(wt+∅) (On differentiating) , v : velocity of block
Now ,
Total energy (E) = Kinetic Energy (KE) + Potential Energy (PE)
PE =
k
[Potential energy of spring]
KE =
m
Now,
k=m
[from 1]
Substituting the value of k,x and v in the equation of KE and PE
We get,
PE =
m

KE =
m
∴ TE =
m
+
m
=
m
Given , TE = 0.163
∴ 0.163 =
m
Substituting m=304g=0.304kg
w = 13.45 
We get,
A = 0.07698 m