A particle executes simple harmonic motion wi | vedclass

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

Question

(A) The displacement of a particle in $SHM$ starting from the equilibrium position is given by $x = A \sin(\omega t)$.Given $A = 10 \,cm$ and $x = 5 \

Vedclass · Accepted Answer

$1 \,m/s$

Vedclass · Answer

(A) The displacement of a particle in $SHM$ starting from the equilibrium position is given by $x = A \sin(\omega t)$.Given $A = 10 \,cm$ and $x = 5 \,cm$,we have $5 = 10 \sin(\omega t)$,which implies $\sin(\omega t) = 1/2$.Thus,$\omega t = \pi/6$. Since $\omega = 2\pi/T$,we get $t = \frac{\pi/6}{2\pi/T} = T/12$.Given $T = 0.6 \,s$,the time interval is $t = 0.6/12 = 0.05 \,s$.The mean velocity $V_{	ext{mean}}$ is defined as the total displacement divided by the total time interval:$V_{	ext{mean}} = \frac{\Delta x}{\Delta t} = \frac{x_f - x_i}{t_f - t_i}$.Starting from equilibrium ($x_i = 0$ at $t_i = 0$) to $x_f = 5 \,cm$ at $t_f = 0.05 \,s$:$V_{	ext{mean}} = \frac{5 \,cm - 0}{0.05 \,s} = 100 \,cm/s = 1 \,m/s$.

$A$ particle executes simple harmonic motion with a time period $0.6 \,s$ and amplitude $10 \,cm$. Find the mean velocity of the particle over the time interval during which it travels a distance $5 \,cm$ starting from the equilibrium position.

Similar Questions

The equations for the displacements of two particles in simple harmonic motion are $y_1=0.1 \sin \left(100 \pi t+\frac{\pi}{3}\right)$ and $y_2=0.1 \cos \pi t$ respectively. The phase difference between the velocities of the two particles at a time $t=0$ is

The displacement of a particle executing $SHM$ is given by $y = 0.25 \sin(200t) \ cm$. The maximum speed of the particle is $......... \ cm \ s^{-1}$.

$A$ particle executes a linear $S.H.M.$ In two of its positions,the velocities are $V_1$ and $V_2$,and the accelerations are $a_1$ and $a_2$ respectively $(0 < a_1 < a_2)$. The distance between the positions is

The maximum velocity of a particle,executing $S$.$H$.$M$. with an amplitude $7 \ mm$ is $4.4 \ ms^{-1}$. The period of oscillation is $\left[\pi=\frac{22}{7}\right]$. (in $s$)

Maximum speed of a particle in simple harmonic motion is $v_{max}.$ Then average speed of a particle in one complete oscillation is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module