(D) The magnitude of acceleration in simple harmonic motion is given by the formula $|a| = \omega^2 y$,where $a$ is acceleration,$\omega$ is angular f

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

(D) The magnitude of acceleration in simple harmonic motion is given by the formula $|a| = \omega^2 y$,where $a$ is acceleration,$\omega$ is angular frequency,and $y$ is displacement.Given $a = 1.0 \, m/s^2$ and $y = 0.01 \, m$,we have:$1.0 = \omega^2 \times 0.01$$\omega^2 = \frac{1.0}{0.01} = 100$$\omega = 10 \, rad/s$The time period $T$ is related to angular frequency by $T = \frac{2\pi}{\omega}$.Substituting the value of $\omega$:$T = \frac{2\pi}{10} = \frac{\pi}{5} \, s$.

Class 11 Physics Oscillations Acceleration and Force of Simple Harmonic Motion

Medium

$A$ body oscillates with a simple harmonic motion having amplitude $0.05 \, m$. At a certain instant,its displacement is $0.01 \, m$ and acceleration is $1.0 \, m/s^2$. The period of oscillation is

AIIMS 2019 All AIIMS

A
$0.1 \, s$
B
$0.2 \, s$
C
$\frac{\pi}{10} \, s$
D
$\frac{\pi}{5} \, s$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

Oscillations

Subtopic

Acceleration and Force of Simple Harmonic Motion

Previous Year

AIIMS 2019 Paper All AIIMS years →

The $x-t$ graph of a particle performing simple harmonic motion is shown in the figure. The acceleration of the particle at $t=2 \ s$ is:

MediumNEET 2023

View Solution

$A$ coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency $\omega$. The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time

MediumAIEEE 2006

View Solution

$A$ particle of mass $1 \ kg$ is moving in $SHM$ with a path length of $0.01 \ m$ and frequency of $50 \ Hz$. The maximum force in newton,acting on the particle is (in $\pi^2$)

MediumMHT CET 2016

View Solution

For a particle performing $S.H.M.$ given by the equation $y = 2 \sin \left( \frac{\pi t}{2} + \phi \right) \, (cm)$,what is the maximum acceleration of the particle?

MediumAIIMS 2014

View Solution

Obtain the formula for the acceleration of a particle from the formula of displacement of $SHM$.

Medium

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

$A$ body oscillates with a simple harmonic motion having amplitude $0.05 \, m$. At a certain instant,its displacement is $0.01 \, m$ and acceleration is $1.0 \, m/s^2$. The period of oscillation is

Similar Questions

The $x-t$ graph of a particle performing simple harmonic motion is shown in the figure. The acceleration of the particle at $t=2 \ s$ is:

$A$ coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency $\omega$. The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time

$A$ particle of mass $1 \ kg$ is moving in $SHM$ with a path length of $0.01 \ m$ and frequency of $50 \ Hz$. The maximum force in newton,acting on the particle is (in $\pi^2$)

For a particle performing $S.H.M.$ given by the equation $y = 2 \sin \left( \frac{\pi t}{2} + \phi \right) \, (cm)$,what is the maximum acceleration of the particle?

Obtain the formula for the acceleration of a particle from the formula of displacement of $SHM$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module