When a mass of
m = 304g
is attached to a spring and the mass-spring system is set into oscillatory motion, the period of the motion is t = 0.467
determine the amplitude (in meters) of the oscillation, if the total energy of the oscillating system is 0.163 j

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 = -- 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

answer: a) y=-8x+5

a) m= -8

b) m= -2

c) m= 7

d) m= 6