A particle performs simple harmonic motion wi | vedclass

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

Question

(B) Let the initial amplitude be $A$ and the new amplitude be $A'$. The velocity $v$ of a particle in simple harmonic motion at a distance $x$ from th

Vedclass · Accepted Answer

$\frac{7A}{3}$

Vedclass · Answer

(B) Let the initial amplitude be $A$ and the new amplitude be $A'$. The velocity $v$ of a particle in simple harmonic motion at a distance $x$ from the equilibrium position is given by $v = \omega \sqrt{A^2 - x^2}$.At $x = \frac{2A}{3}$,the initial velocity $v$ is $v = \omega \sqrt{A^2 - (\frac{2A}{3})^2} = \omega \sqrt{A^2 - \frac{4A^2}{9}} = \omega \sqrt{\frac{5A^2}{9}} = \frac{\omega A \sqrt{5}}{3}$.The new speed is $v' = 3v = \omega A \sqrt{5}$.Using the formula for the new amplitude $A'$ at the same position $x = \frac{2A}{3}$:$(v')^2 = \omega^2 (A'^2 - x^2)$$(\omega A \sqrt{5})^2 = \omega^2 (A'^2 - (\frac{2A}{3})^2)$$5A^2 = A'^2 - \frac{4A^2}{9}$$A'^2 = 5A^2 + \frac{4A^2}{9} = \frac{45A^2 + 4A^2}{9} = \frac{49A^2}{9}$$A' = \sqrt{\frac{49A^2}{9}} = \frac{7A}{3}$.

$A$ particle performs simple harmonic motion with amplitude $A$. Its speed is trebled at the instant that it is at a distance $\frac{2A}{3}$ from the equilibrium position. The new amplitude of the motion is

Similar Questions

$A$ particle is making simple harmonic motion along the $x$-axis. If at a distance $x_{1}$ and $x_{2}$ from the mean position the velocities of the particle are $v_{1}$ and $v_{2}$ respectively,the time period of its oscillation is given as -

$A$ simple harmonic oscillator has a period of $0.01 \, s$ and an amplitude of $0.2 \, m$. The magnitude of the velocity in $m \cdot s^{-1}$ at the centre of oscillation is:

$A$ particle executing simple harmonic motion starts from the mean position with amplitude $A$ and periodic time $T$. At what displacement is its speed one-fourth of the maximum speed?

The instantaneous displacement of a particle in $S.H.M.$ is $x = A \cos \left(\omega t + \frac{\pi}{4}\right)$. The time at which the velocity is maximum for the first time is

$A$ 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

Vedclass Test Series

Exam Paper Generator

Online Exam Module