JEE MAIN - Physics (2014 (Offline) - No. 18)
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:
amplitude of motion is $$3a$$
time period of oscillations is $$8\tau $$
amplitude of motion is $$4a$$
time period of oscillations is $$6\tau $$
Explanation
In simple harmonic motion, starting from rest,
At $$t=0,$$ $$x=A$$
$$x = A\cos \omega t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
When $$t = \tau ,\,\,x = A - a$$
When $$t = 2\,\tau ,\,x = A - 3a$$
From equation $$(i)$$
$$A - a = A\cos \omega \,\tau \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
$$A - 3a = A\cos 2\omega \,\tau \,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {iii} \right)$$
As $$\cos 2\omega \,\tau = 2{\cos ^2}\omega \tau - 1...\left( {iv} \right)$$
From equation $$(ii),$$ $$(iii)$$ and $$(iv)$$
$${{A - 3A} \over A} = 2{\left( {{{A - a} \over A}} \right)^2} - 1$$
$$ \Rightarrow {{A - 3a} \over A} = {{2{A^2} + 2{a^2} - 4Aa - {A^2}} \over {{A^2}}}$$
$$ \Rightarrow {A^2} - 3aA = {A^2} + 2{a^2} - 4Aa$$
$$ \Rightarrow 2{a^2} = aA \Rightarrow \,\,\,\,\,\,\,A = 2a$$
$$ \Rightarrow {a \over A} = {1 \over 2}$$
Now, $$A-a=A$$ $$\cos \omega \tau $$
$$ \Rightarrow \cos \omega \tau = {{A - a} \over A} \Rightarrow \,\,\cos \omega \tau = {1 \over 2}$$
or, $${{2\pi } \over T}\tau = {\pi \over 3} \Rightarrow \,\,\,T - 6\,\tau $$
At $$t=0,$$ $$x=A$$
$$x = A\cos \omega t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
When $$t = \tau ,\,\,x = A - a$$
When $$t = 2\,\tau ,\,x = A - 3a$$
From equation $$(i)$$
$$A - a = A\cos \omega \,\tau \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
$$A - 3a = A\cos 2\omega \,\tau \,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {iii} \right)$$
As $$\cos 2\omega \,\tau = 2{\cos ^2}\omega \tau - 1...\left( {iv} \right)$$
From equation $$(ii),$$ $$(iii)$$ and $$(iv)$$
$${{A - 3A} \over A} = 2{\left( {{{A - a} \over A}} \right)^2} - 1$$
$$ \Rightarrow {{A - 3a} \over A} = {{2{A^2} + 2{a^2} - 4Aa - {A^2}} \over {{A^2}}}$$
$$ \Rightarrow {A^2} - 3aA = {A^2} + 2{a^2} - 4Aa$$
$$ \Rightarrow 2{a^2} = aA \Rightarrow \,\,\,\,\,\,\,A = 2a$$
$$ \Rightarrow {a \over A} = {1 \over 2}$$
Now, $$A-a=A$$ $$\cos \omega \tau $$
$$ \Rightarrow \cos \omega \tau = {{A - a} \over A} \Rightarrow \,\,\cos \omega \tau = {1 \over 2}$$
or, $${{2\pi } \over T}\tau = {\pi \over 3} \Rightarrow \,\,\,T - 6\,\tau $$
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