A spring executes $SHM$ with mass of $10\,kg$ attached to it. The force constant of spring is $10\,N/m$.If at any instant its velocity is $40\,cm/sec$, the displacement will be .... $m$ (where amplitude is $0.5\,m$)
Medium
Download our app for free and get started
(b) Angular velocity $\omega = \sqrt {\left( {\frac{k}{m}} \right)} $$ = \sqrt {\left( {\frac{{10}}{{10}}} \right)} = 1$
Now $u = \omega \sqrt {{a^2} - {y^2}} $
Now $u = \omega \sqrt {{a^2} - {y^2}} $
==>${y^2} = {a^2} - \frac{{{u^2}}}{{{\omega ^2}}}$$ = {(0.5)^2} - \frac{{{{(0.4)}^2}}}{{{1^2}}}$
==> ${y^2} = 0.9$
$y = 0.3\,m$
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1A pendulum is formed by pivoting a disc. What distance from center of mass, it should be pivoted for minimum time period while performing $SHM$ ?View Solution
- 2A mass $m$ is suspended separately by two different springs of spring constant $K_1$ and $K_2$ gives the time-period ${t_1}$ and ${t_2}$ respectively. If same mass $m$ is connected by both springs as shown in figure then time-period $t$ is given by the relationView Solution
- 3A mass $m$ is attached to two springs as shown in figure. The spring constants of two springs are $K _1$ and $K _2$. For the frictionless surface, the time period of oscillation of mass $m$ isView Solution
- 4A particle performs $S.H.M.$ of amplitude $A$ with angular frequency $\omega$ along a straight line. Whenit is at a distance $\frac{{\sqrt 3 }}{2}$ $A$ from mean position, its kinetic energy gets increased by an amount $\frac{1}{2}m{\omega ^2}{A^2}$ due to an impulsive force. Then its new amplitude becomesView Solution
- 5View SolutionAn assembly of identical spring-mass systems is placed on a smooth horizontal surface as shown. Initially the springs are relaxed. The left mass is displaced to the left while the right mass is displaced to the right and released. The resulting collision is elastic. The time period of the oscillations of the system is :-
- 6View SolutionIn forced oscillations, a particle oscillates simple harmonically with frequency equal to
- 7A vibratory motion is represented by $x = 2A\,\cos \omega t + A\,\cos \,\left( {\omega t + \frac{\pi }{2}} \right) + A\,\cos \,\left( {\omega t + \pi } \right)$ $ + \frac{A}{2}\,\cos \left( {\omega t + \frac{{3\pi }}{2}} \right)$. The resultant amplitude of the motion isView Solution
- 8The metallic bob of simple pendulum has the relative density $5$. The time period of this pendulum is $10\,s$. If the metallic bob is immersed in water, then the new time period becomes $5 \sqrt{ x } s$. The value of $x$ will be.View Solution
- 9An object of mass $0.2\, kg$ executes simple harmonic along $X-$ axis with frequency of $\frac{{25}}{\pi }Hz$. At the position $x = 0.04m$, the object has kinetic energy of $0.5 \,J$ and potential energy of $0.4\, J$ amplitude of oscillation in meter is equal toView Solution
- 10Two particles $P$ and $Q$ describe simple harmonic motions of same period, same amplitude, along the same line about the same equilibrium position $O.$ When $P$ and $Q$ are on opposite sides of $O$ at the same distance from $O$ they have the same speed of $1.2 \,m/s$ in the same direction, when their displacements are the same they have the same speed of $1.6\, m/s$ in opposite directions. The maximum velocity in $m/s$ of either particle isView Solution