(C) The velocity of a particle in simple harmonic motion at the mean position is given by $V_{max} = A\omega$.Given $A = 4 \ cm$ and $V_{max} = 10 \ c

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

(C) The velocity of a particle in simple harmonic motion at the mean position is given by $V_{max} = A\omega$.Given $A = 4 \ cm$ and $V_{max} = 10 \ cm/s$,we have $10 = 4\omega$,which gives $\omega = 2.5 \ rad/s$.The velocity $V$ at any displacement $x$ from the mean position is given by $V = \omega \sqrt{A^2 - x^2}$.Substituting the given values: $5 = 2.5 \sqrt{4^2 - x^2}$.Dividing by $2.5$: $2 = \sqrt{16 - x^2}$.Squaring both sides: $4 = 16 - x^2$.Therefore,$x^2 = 12$,which means $x = \sqrt{12} \ cm$.Comparing this with $\sqrt{\alpha} \ cm$,we get $\alpha = 12$.

Class 11 Physics Oscillations Velocity of Simple Harmonic Motion

Difficult

$A$ particle executes simple harmonic motion with an amplitude of $4 \ cm$. At the mean position,the velocity of the particle is $10 \ cm/s$. The distance of the particle from the mean position when its speed becomes $5 \ cm/s$ is $\sqrt{\alpha} \ cm$,where $\alpha = $ . . . . . . .

JEE MAIN 2024 All JEE MAIN

A
$11$
B
$22$
C
$12$
D
$15$

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

JEE MAIN 2024 Paper All JEE MAIN years →

If a particle under $S.H.M.$ has a time period of $0.1 \ s$ and an amplitude of $2 \times 10^{-3} \ m$,what is its maximum velocity?

Easy

View Solution

$A$ point performs simple harmonic oscillation of period $T$ and the equation of motion is given by $x = A \sin(\omega t + \frac{\pi}{6})$. After the elapse of what fraction of the time period will the velocity of the point be equal to half of its maximum velocity?

MediumAIPMT 2008

View Solution

The velocity vector $v$ and displacement vector $x$ of a particle executing $SHM$ are related as $\frac{v dv}{dx} = -\omega^2 x$ with the initial condition $v = v_0$ at $x = 0$. The velocity $v$,when displacement is $x$,is

DifficultAIIMS 2015

View Solution

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

DifficultTS EAMCET 2001

View Solution

The maximum velocity of a particle performing $S.H.M.$ is $V$. If the periodic time is made $\left(\frac{1}{3}\right)^{rd}$ of its original value and the amplitude is doubled,then the new maximum velocity of the particle will be:

MediumMHT CET 2023

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 executes simple harmonic motion with an amplitude of $4 \ cm$. At the mean position,the velocity of the particle is $10 \ cm/s$. The distance of the particle from the mean position when its speed becomes $5 \ cm/s$ is $\sqrt{\alpha} \ cm$,where $\alpha = $ . . . . . . .

Similar Questions

If a particle under $S.H.M.$ has a time period of $0.1 \ s$ and an amplitude of $2 \times 10^{-3} \ m$,what is its maximum velocity?

$A$ point performs simple harmonic oscillation of period $T$ and the equation of motion is given by $x = A \sin(\omega t + \frac{\pi}{6})$. After the elapse of what fraction of the time period will the velocity of the point be equal to half of its maximum velocity?

The velocity vector $v$ and displacement vector $x$ of a particle executing $SHM$ are related as $\frac{v dv}{dx} = -\omega^2 x$ with the initial condition $v = v_0$ at $x = 0$. The velocity $v$,when displacement is $x$,is

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 maximum velocity of a particle performing $S.H.M.$ is $V$. If the periodic time is made $\left(\frac{1}{3}\right)^{rd}$ of its original value and the amplitude is doubled,then the new maximum velocity of the particle will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module