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(A) The displacement of a particle in $S.H.M.$ starting from the mean position is given by $x(t) = A \sin(\omega t)$,where $A$ is the amplitude and $\

Vedclass · Accepted Answer

$1:(\sqrt{2}-1)$

Vedclass · Answer

(A) The displacement of a particle in $S.H.M.$ starting from the mean position is given by $x(t) = A \sin(\omega t)$,where $A$ is the amplitude and $\omega = 2\pi/T$. Given $T = 8 \ s$,we have $\omega = 2\pi/8 = \pi/4 \ rad/s$.At $t=0$,$x(0) = 0$.At $t=1 \ s$,$x(1) = A \sin(\pi/4 	imes 1) = A/\sqrt{2}$. Distance in $1^{st}$ second $d_1 = |x(1) - x(0)| = A/\sqrt{2}$.At $t=2 \ s$,$x(2) = A \sin(\pi/4 	imes 2) = A \sin(\pi/2) = A$. Distance in $2^{nd}$ second $d_2 = |x(2) - x(1)| = |A - A/\sqrt{2}| = A(1 - 1/\sqrt{2}) = A(\sqrt{2}-1)/\sqrt{2}$.The ratio $d_1/d_2 = (A/\sqrt{2}) / [A(\sqrt{2}-1)/\sqrt{2}] = 1/(\sqrt{2}-1)$.

$A$ particle performing linear $S.H.M.$ has a period of $8 \ s$. At time $t=0$,it is at the mean position. The ratio of the distances travelled by the particle in the $1^{st}$ and $2^{nd}$ second is $(\cos 45^{\circ} = 1/\sqrt{2})$.

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