A particle vibrating simple harmonically has | vedclass

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

Question

(C) In Simple Harmonic Motion $(SHM)$, the acceleration $a$ of a vibrating particle is given by the relation $a = \omega^2 x$, where $\omega$ is the a

Vedclass · Accepted Answer

$3.142$

Vedclass · Answer

(C) In Simple Harmonic Motion $(SHM)$, the acceleration $a$ of a vibrating particle is given by the relation $a = \omega^2 x$, where $\omega$ is the angular frequency and $x$ is the displacement from the mean position.Given: Acceleration $a = 16 \ cm/s^2$ and displacement $x = 4 \ cm$.Substituting these values into the formula:$16 = \omega^2 	imes 4$$\omega^2 = \frac{16}{4} = 4$$\omega = 2 \ rad/s$The time period $T$ is related to angular frequency by the formula $T = \frac{2\pi}{\omega}$.$T = \frac{2\pi}{2} = \pi \ s$Using $\pi \approx 3.142$, we get $T = 3.142 \ s$.

$A$ particle vibrating simple harmonically has an acceleration of $16 \ cm/s^2$ when it is at a distance of $4 \ cm$ from the mean position. Its time period is: (in $s$)

Similar Questions

In $S.H.M.$,maximum acceleration is at

The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is $10\,s^{-1}$. At $t = 0$,the displacement is $5\,m$. What is the maximum acceleration? The initial phase is $\frac{\pi}{4}$.

Obtain the force law for $SHM$ from the displacement of $SHM$ particle.

The magnitude of maximum acceleration is $\pi$ times that of maximum velocity of a simple harmonic oscillator. The time period of the oscillator in seconds is

$A$ particle of mass $10 \text{ g}$ is executing simple harmonic motion with an amplitude of $0.5 \text{ m}$ and a periodic time of $(\pi / 5) \text{ s}$. The maximum value of the force acting on the particle is ... $N$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module