Question

When a mass of

Answer 1

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 = $\sqrt{\dfrac{k}{m} }$ -- 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 = $\dfrac{2\pi }{w}$ = 0.467

∴ w = $\dfrac{2\pi }{T}$ = $\dfrac{2\pi }{0.467}$ = 13.45 $s^{-1}$

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 = $\frac{1}{2}$k$x^{2}$ [Potential energy of spring]

KE =  $\frac{1}{2}$m$v^{2}$

Now,

k=m$w^{2}$ [from 1]

Substituting the value of k,x and v in the equation of KE and PE

We get,

PE =  $\frac{1}{2}$m$w^{2}$$A^{2} sin^{2}(wt+∅)$

KE =  $\frac{1}{2}$m$A^{2}w^{2}cos^{2}(wt+∅)$

∴ TE = $\frac{1}{2}$m$w^{2}$$A^{2} sin^{2}(wt+∅)$ +     $\frac{1}{2}$m$A^{2}w^{2}cos^{2}(w+∅)$

        = $\frac{1}{2}$m$w^{2}A^{2}$

Given , TE = 0.163

∴ 0.163 = $\frac{1}{2}$m$w^{2}A^{2}$

Substituting m=304g=0.304kg

                    w = 13.45 $s^{-1}$

We get,

A = 0.07698 m