Question 242: Oscillations

Question

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

Accepted Answer

Step 1: Understand the motion
Time period: $T = 8 \, 	ext{s}$.
Therefore, angular frequency:
$ \omega = \frac{2\pi}{T} = \frac{2\pi}{8} = \frac{\pi}{4} \, 	ext{rad/s}. $
At $t=0$, the particle is at the mean position.
So displacement function:
$ x(t) = A \sin(\omega t). $
Velocity: $v(t) = \frac{dx}{dt} = A\omega \cos(\omega t).$
Step 2: Distance travelled in 1st second (0 to 1s)
Distance travelled = integral of $|v|$.
For $t = 0$ to small values, $\cos(\omega t) > 0$ (since $\omega = \pi/4$, in 1 second $\omega t = \pi/4 \approx 45^\circ$, still positive). So velocity doesnÃ¢ÂÂt change sign in this interval Ã¢ÂÂ monotonic increase of $x$.
Hence distance travelled = displacement magnitude:
$ s_1 = x(1) - x(0) = A\sin\left(\frac{\pi}{4}\cdot 1ight) - 0 = A\sin\frac{\pi}{4} = \frac{A}{\sqrt{2}}. $
Step 3: Distance travelled in 2nd second (1 to 2s)
Check velocity sign between $t=1$s and $t=2$s.
At $t=1$, $\omega t = \pi/4$, cos positive. At $t=2$, $\omega t = \pi/2$, cos zero. So velocity gradually reduces to zero without changing sign. So again monotonic increase in displacement.
Thus distance = displacement change:
$ s_2 = x(2) - x(1). $
$ x(2) = A \sin\left( \frac{\pi}{4} \cdot 2 ight) = A\sin\frac{\pi}{2} = A. $
So
$ s_2 = A - \frac{A}{\sqrt{2}} = A\left(1 - \frac{1}{\sqrt{2}}ight). $
Step 4: Ratio $s_1 : s_2$
$ \frac{s_1}{s_2} = \frac{	frac{A}{\sqrt{2}}}{A\left(1 - 	frac{1}{\sqrt{2}}ight)}  = \frac{\frac{1}{\sqrt{2}}}{1 - \frac{1}{\sqrt{2}}}. $
Simplify denominator: $1 - 	frac{1}{\sqrt{2}} = \frac{\sqrt{2}-1}{\sqrt{2}}$.
So:
$ \frac{s_1}{s_2} = \frac{	frac{1}{\sqrt{2}}}{	frac{\sqrt{2}-1}{\sqrt{2}}} = \frac{1}{\sqrt{2} - 1}. $
Multiply numerator and denominator:
$\frac{1}{\sqrt{2} - 1} = \sqrt{2}+1.$
Thus ratio:
$ s_1 : s_2 = (\sqrt{2}+1) : 1. $
Step 5: Match with options
They ask ratio of distances in 1st and 2nd second = $1 : (\sqrt{2}-1)$ or $2:1$ etc.
We got $(\sqrt{2}+1):1$.
So equivalently $1 : (\sqrt{2}-1)$? LetÃ¢ÂÂs check:
$(\sqrt{2}+1):1$ = multiply numerator and denominator by $(\sqrt{2}-1)$:
$ (\sqrt{2}+1):1 = 1:(\sqrt{2}-1). $
Yes, same.
Ã¢ÂÂ Final Answer:
$ \boxed{1 : (\sqrt{2} - 1)} $
Option A

Answer

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

Answer

$1: 2$

Answer

$2: 1$

Answer

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

Oscillations

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

Choose the correct answer: