A particle executing SHM of amplitude 4 , cm | vedclass

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

Question

(B) The displacement equation for $SHM$ is given by $x = A \sin(\omega t + \phi)$. Assuming the particle starts from the mean position,$x = A \sin(\om

Vedclass · Accepted Answer

$1/3 \, s$

Vedclass · Answer

(B) The displacement equation for $SHM$ is given by $x = A \sin(\omega t + \phi)$. Assuming the particle starts from the mean position,$x = A \sin(\omega t)$.Given amplitude $A = 4 \, cm$ and time period $T = 4 \, s$,the angular frequency is $\omega = \frac{2\pi}{T} = \frac{2\pi}{4} = \frac{\pi}{2} \, rad/s$.For position $x_1 = 2 \, cm$:$2 = 4 \sin(\omega t_1) \Rightarrow \sin(\omega t_1) = 1/2 \Rightarrow \omega t_1 = \pi/6$.For position $x_2 = 2\sqrt{3} \, cm$:$2\sqrt{3} = 4 \sin(\omega t_2) \Rightarrow \sin(\omega t_2) = \sqrt{3}/2 \Rightarrow \omega t_2 = \pi/3$.The time taken $\Delta t = t_2 - t_1$ is given by:$\omega(t_2 - t_1) = \pi/3 - \pi/6 = \pi/6$.Substituting $\omega = \pi/2$:$(\pi/2) \Delta t = \pi/6 \Rightarrow \Delta t = \frac{\pi}{6} 	imes \frac{2}{\pi} = 1/3 \, s$.

$A$ particle executing $SHM$ of amplitude $4 \, cm$ and $T = 4 \, s$. The time taken by it to move from $+2 \, cm$ to $+2\sqrt{3} \, cm$ is

Similar Questions

$A$ 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.

The displacement versus time curve for a particle executing $SHM$ is shown in the figure. Identify the points marked at which $(i)$ the velocity of the oscillator is zero,$(ii)$ the speed of the oscillator is maximum.

$A$ particle performs simple harmonic motion between $x = -A$ and $x = +A$. How much time does it take to travel from $x = A$ to $x = A/2$?

Two particles $A$ and $B$ execute $SHMs$ of periods $T$ and $\frac{3T}{2}$. If they start from the mean position,then the phase difference between them,when the particle $A$ completes two oscillations,will be

$A$ particle executes simple harmonic motion between $x=-A$ and $x=+A$. If it takes a time $T_1$ to go from $x=0$ to $x=A/2$ and $T_2$ to go from $x=A/2$ to $x=A$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module