A particle oscillates in a straight line simp | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The equation for the displacement of a particle starting from the mean position in $SHM$ is given by $x(t) = A \sin(\omega t)$.Given: $T = 8 \ s$,

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The equation for the displacement of a particle starting from the mean position in $SHM$ is given by $x(t) = A \sin(\omega t)$.Given: $T = 8 \ s$,so $\omega = \frac{2\pi}{T} = \frac{2\pi}{8} = \frac{\pi}{4} \ rad/s$. Amplitude $A = 4\sqrt{2} \ m$.Thus,$x(t) = 4\sqrt{2} \sin(\frac{\pi}{4} t)$.Distance in the $1^{	ext{st}}$ second ($t=0$ to $t=1$): $x(1) = 4\sqrt{2} \sin(\frac{\pi}{4}) = 4\sqrt{2} 	imes \frac{1}{\sqrt{2}} = 4 \ m$.Distance in the $2^{	ext{nd}}$ second ($t=1$ to $t=2$): $x(2) = 4\sqrt{2} \sin(\frac{2\pi}{4}) = 4\sqrt{2} \sin(\frac{\pi}{2}) = 4\sqrt{2} 	imes 1 = 4\sqrt{2} \ m$.The distance travelled in the $2^{	ext{nd}}$ second is $d_2 = x(2) - x(1) = 4\sqrt{2} - 4 = 4(\sqrt{2} - 1) \ m$.The ratio of distance in the $1^{	ext{st}}$ second to the $2^{	ext{nd}}$ second is $\frac{d_1}{d_2} = \frac{4}{4(\sqrt{2} - 1)} = \frac{1}{\sqrt{2} - 1}$.

Similar Questions

$A$ particle executes $SHM$ with a period of $1.2 \, s$ and an amplitude of $8 \, cm$. Find the time it takes to travel $3 \, cm$ from the positive extremity of its oscillation.

$A$ particle is performing simple harmonic motion with an amplitude of $4 \, cm$ and a time period of $12 \, s$. What is the ratio of the time taken by the particle to travel from its mean position to $2 \, cm$ to the time taken to travel from $2 \, cm$ to its extreme position?

$A$ particle executes $S.H.M.$ of amplitude $A$ along the $x$-axis. At $t = 0$,the position of the particle is $x = \frac{A}{2}$ and it moves along the positive $x$-axis. If the displacement of the particle in time $t$ is $x = A \sin (\omega t + \delta)$,then the value of $\delta$ will be:

What is the phase difference between the simple harmonic motions $x = a \sin(\omega t - \alpha)$ and $y = b \cos(\omega t - \alpha)$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module