Question 150: 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:

Choose the correct answer: