Question 253: Oscillations

Question

A particle oscillates in straight line simple harmonically with period 8 second and amplitude $4 \sqrt{2} \mathrm{~m}$. Particle starts from mean position. The ratio of the distance travelled by it in $1^{	ext {st }}$ second of its motion to that in $2^{	ext {nd }}$ second is $\left(\sin 45^{\circ}=1 / \sqrt{2}, \sin \frac{\pi}{2}=1ight)$

Accepted Answer

For a particle executing SHM from mean position,
$$ \begin{array}{rl} x & =A \sin \omega t \ A t & t=1 s \ x_1 & =A \sin \left(\frac{2 \pi}{T} tight)=A \sin \left(\frac{2 \pi}{8} 	imes 1ight) \ & =A \sin 45^{\circ}=\frac{A}{\sqrt{2}} \ t & =2 s \ x_2 & =A \sin \left(\frac{2 \pi}{T} tight)=A \sin \left(\frac{2 \pi}{8} 	imes 2ight) \ & =A \sin 90^{\circ}=A \end{array} $$
$	herefore \quad$ Distance covered in $2^{	ext {nd }}$ second,
$$ x=x_2-x_1=A-\frac{A}{\sqrt{2}}=A\left(\frac{\sqrt{2}-1}{\sqrt{2}}ight) $$
$	herefore \quad$ Required ratio,
$$ \frac{x_1}{x}=\frac{A / \sqrt{2}}{A\left(\frac{\sqrt{2}-1}{\sqrt{2}}ight)}=\frac{4 \sqrt{2} / \sqrt{2}}{4 \sqrt{2}\left(\frac{\sqrt{2}-1}{\sqrt{2}}ight)}=\frac{1}{\sqrt{2}-1} $$

Answer

$1: 8$

Answer

$1: 4$

Answer

$1: 2$

Answer

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

Oscillations

Choose the correct answer: