Answer:
The amplitude of oscillation is 0.07698m
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 = [tex]\sqrt{\dfrac{k}{m} }[/tex] -- 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 = [tex]\dfrac{2\pi }{w}[/tex] = 0.467
∴ w = [tex]\dfrac{2\pi }{T}[/tex] = [tex]\dfrac{2\pi }{0.467}[/tex] = 13.45 [tex]s^{-1}[/tex]
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 = [tex]\frac{1}{2}[/tex]k[tex]x^{2}[/tex] [Potential energy of spring]
KE = [tex]\frac{1}{2}[/tex]m[tex]v^{2}[/tex]
Now,
k=m[tex]w^{2}[/tex] [from 1]
Substituting the value of k,x and v in the equation of KE and PE
We get,
PE = [tex]\frac{1}{2}[/tex]m[tex]w^{2}[/tex][tex]A^{2} sin^{2}(wt+∅)[/tex]
KE = [tex]\frac{1}{2}[/tex]m[tex]A^{2}w^{2}cos^{2}(wt+∅)[/tex]
∴ TE = [tex]\frac{1}{2}[/tex]m[tex]w^{2}[/tex][tex]A^{2} sin^{2}(wt+∅)[/tex] + [tex]\frac{1}{2}[/tex]m[tex]A^{2}w^{2}cos^{2}(w+∅)[/tex]
= [tex]\frac{1}{2}[/tex]m[tex]w^{2}A^{2}[/tex]
Given , TE = 0.163
∴ 0.163 = [tex]\frac{1}{2}[/tex]m[tex]w^{2}A^{2}[/tex]
Substituting m=304g=0.304kg
w = 13.45 [tex]s^{-1}[/tex]
We get,
A = 0.07698 m
