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A partical moves simple harmonically in a straight line starting from rest. In first $\mathrm{t}$ second it travels $\mathrm{a}$ distance $\mathrm{' a '}$ and in hent $\mathrm{t}$ second it travels a distance $\mathrm{' 2 c '}$ in the same direction, time period will be -

Option: 1
$\mathrm{4\: t}$

Option: 2
$\mathrm{5\: t}$

Option: 3
$\mathrm{6\: t}$

Option: 4
$\mathrm{8\: t}$

Answers (1)

best_answer

amplitude $\mathrm{A =\left[\frac{2 a^2}{3 a-b}\right] }$

$\mathrm{A =\frac{2 a^2}{3 a-2 a}=2 a}$

As particle starts from rest, so

$\mathrm{x =A \cos \omega t }$

$\mathrm{ =2 a \cos \omega t}$

$\mathrm{ \text { or } A-a=2 a \cos \left(\frac{2 \pi t}{T}\right) }$

$\mathrm{ 2 a-a=2 a \cos \frac{2 \pi t}{T} }$

$\mathrm{ \text { or } \cos \left(\frac{2 \pi t}{T}\right)=\frac{1}{2}}$

$\mathrm{\frac{2\: \pi\: t}{T}=\frac{\pi}{3}}$

$\mathrm{T=6t}$

Hence option 3 is correct.

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Riya

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