A particle executes simple harmonic motion be | vedclass

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

Question

(D) The displacement of a particle in simple harmonic motion is given by $x = A \sin(\omega t)$.For the particle to go from $x = 0$ to $x = A/2$ in ti

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) The displacement of a particle in simple harmonic motion is given by $x = A \sin(\omega t)$.For the particle to go from $x = 0$ to $x = A/2$ in time $t_1 = 2 \, s$:$A/2 = A \sin(\omega t_1) \implies \sin(\omega t_1) = 1/2 \implies \omega t_1 = \pi/6$.Thus,$\omega(2) = \pi/6 \implies \omega = \pi/12 \, rad/s$.For the particle to go from $x = A/2$ to $x = A$ in time $t_2$:At $x = A/2$,the phase is $\phi_1 = \pi/6$.At $x = A$,the phase is $\phi_2 = \pi/2$.The phase difference is $\Delta \phi = \phi_2 - \phi_1 = \pi/2 - \pi/6 = \pi/3$.Since $\Delta \phi = \omega t_2$,we have $\pi/3 = (\pi/12) t_2$.Solving for $t_2$,we get $t_2 = (\pi/3) 	imes (12/\pi) = 4 \, s$.

$A$ particle executes simple harmonic motion between $x = -A$ and $x = +A$. If the time taken by the particle to go from $x = 0$ to $x = A/2$ is $2 \, s$,then the time taken by the particle in going from $x = A/2$ to $x = A$ is $......... \, s$.

Similar Questions

The displacement of a particle undergoing $SHM$ with time period $T$ is given by $x(t) = x_m \cos(\omega t + \phi)$. The particle is at $x = -x_m$ at time $t = 0$. The particle is at $x = +x_m$ when:

The displacements of two particles executing $S.H.M.$ on the same line are given as $y_1 = a \sin \left(\frac{\pi}{2} t + \phi\right)$ and $y_2 = b \sin \left(\frac{2 \pi}{3} t + \phi\right)$. The phase difference between them at $t = 1 \, s$ is .........

$A$ particle executes simple harmonic oscillation with an amplitude $a$. The period of oscillation is $T$. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is:

Time period of a particle executing $SHM$ is $8 \, s$. At $t = 0$ it is at the mean position. The ratio of the distance covered by the particle in the $1^{st}$ second to the $2^{nd}$ second is:

$A$ particle is executing simple harmonic motion. Its amplitude is $A$ and time period is $5 \text{ sec}$. The time required by it to move from $x = A$ to $x = A/\sqrt{2}$ is . . . . . . sec.

Vedclass Test Series

Exam Paper Generator

Online Exam Module