Question 126: Oscillations

Question

A particle moves with simple harmonic motion in a straight line. In first $	au s,$ after starting from rest it travels a distance $a,$ and in next $	au s$ it travels $2a,$ in same direction, then:

Accepted Answer

In simple harmonic motion, starting from rest,
At $t=0,$ $x=A$
$x = A\cos \omega t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i ight)$
When $t = 	au ,\,\,x = A - a$
When $t = 2\,	au ,\,x = A - 3a$
From equation $(i)$
$A - a = A\cos \omega \,	au \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} ight)$
$A - 3a = A\cos 2\omega \,	au \,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {iii} ight)$
As $\cos 2\omega \,	au  = 2{\cos ^2}\omega 	au  - 1...\left( {iv} ight)$
From equation $(ii),$ $(iii)$ and $(iv)$
${{A - 3A} \over A} = 2{\left( {{{A - a} \over A}} ight)^2} - 1$
$ \Rightarrow {{A - 3a} \over A} = {{2{A^2} + 2{a^2} - 4Aa - {A^2}} \over {{A^2}}}$
$ \Rightarrow {A^2} - 3aA = {A^2} + 2{a^2} - 4Aa$
$ \Rightarrow 2{a^2} = aA \Rightarrow \,\,\,\,\,\,\,A = 2a$
$ \Rightarrow {a \over A} = {1 \over 2}$
Now, $A-a=A$ $\cos \omega 	au $
$ \Rightarrow \cos \omega 	au  = {{A - a} \over A} \Rightarrow \,\,\cos \omega 	au  = {1 \over 2}$
or, ${{2\pi } \over T}	au  = {\pi  \over 3} \Rightarrow \,\,\,T - 6\,	au $

Answer

amplitude of motion  is $3a$

Answer

time period of oscillations is $8	au $

Answer

amplitude of motion is $4a$

Answer

time period of oscillations is $6	au $

Oscillations

A particle moves with simple harmonic motion in a straight line. In first \tau s, after starting from rest it travels a distance a, and in next \tau s it travels 2a, in same direction, then:

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