(D) Given, amplitude, $A = 0.2 \,cm = 2 \times 10^{-3} \,m$. Velocity at the mean position in simple harmonic motion is the maximum velocity, $v_{\tex

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

(D) Given, amplitude, $A = 0.2 \,cm = 2 \times 10^{-3} \,m$. Velocity at the mean position in simple harmonic motion is the maximum velocity, $v_{\text{max}} = 5 \,m/s$. We know that the maximum velocity is given by the formula, $v_{\text{max}} = A \omega$. Therefore, the angular frequency $\omega$ is given by, $\omega = \frac{v_{\text{max}}}{A}$. Substituting the values, $\omega = \frac{5}{2 \times 10^{-3}} = 2.5 \times 10^3 = 2500 \,rad/s$.

Class 11 Physics Oscillations Velocity of Simple Harmonic Motion

Easy

$A$ particle is executing simple harmonic motion in one-dimension. If the amplitude of oscillations is $0.2 \,cm$ and if its velocity at the mean position is $5 \,m/s$, then the angular frequency of the oscillation is

TS EAMCET 2020 All TS EAMCET

A
$1000 \,rad/s$
B
$1500 \,rad/s$
C
$2000 \,rad/s$
D
$2500 \,rad/s$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

Oscillations

Subtopic

Velocity of Simple Harmonic Motion

Previous Year

TS EAMCET 2020 Paper All TS EAMCET years →

Two particles $P$ and $Q$ start from the origin and execute simple harmonic motion along the $X$-axis with the same amplitude but with periods $3 \ s$ and $6 \ s$, respectively. The ratio of the velocities of $P$ and $Q$ when they meet is

DifficultAP EAMCET 2001

View Solution

The position,velocity,and acceleration of a particle executing simple harmonic motion are found to have magnitudes of $4 \ m$,$2 \ ms^{-1}$,and $16 \ ms^{-2}$ at a certain instant. The amplitude of the motion is $\sqrt{x} \ m$,where $x$ is . . . . . .

DifficultJEE MAIN 2024

View Solution

If a simple pendulum oscillates with an amplitude of $50 \,mm$ and time period of $2 \,s$, then its maximum velocity is (in $\,ms^{-1}$)

EasyMHT CET 2009

View Solution

If a simple pendulum oscillates with an amplitude of $50\, mm$ and time period of $2\, s$,then its maximum velocity is .... $m/s$.

EasyAIIMS 1998

View Solution

$A$ particle is performing $S.H.M.$ about $x = 0$, with an amplitude $a$ and time period $T$. The speed of the particle at $x = \frac{a}{3}$ will be

DifficultMHT CET 2026

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

$A$ particle is executing simple harmonic motion in one-dimension. If the amplitude of oscillations is $0.2 \,cm$ and if its velocity at the mean position is $5 \,m/s$, then the angular frequency of the oscillation is

Similar Questions

Two particles $P$ and $Q$ start from the origin and execute simple harmonic motion along the $X$-axis with the same amplitude but with periods $3 \ s$ and $6 \ s$, respectively. The ratio of the velocities of $P$ and $Q$ when they meet is

The position,velocity,and acceleration of a particle executing simple harmonic motion are found to have magnitudes of $4 \ m$,$2 \ ms^{-1}$,and $16 \ ms^{-2}$ at a certain instant. The amplitude of the motion is $\sqrt{x} \ m$,where $x$ is . . . . . .

If a simple pendulum oscillates with an amplitude of $50 \,mm$ and time period of $2 \,s$, then its maximum velocity is (in $\,ms^{-1}$)

If a simple pendulum oscillates with an amplitude of $50\, mm$ and time period of $2\, s$,then its maximum velocity is .... $m/s$.

$A$ particle is performing $S.H.M.$ about $x = 0$, with an amplitude $a$ and time period $T$. The speed of the particle at $x = \frac{a}{3}$ will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module