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