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
Medium
Download our app for free and get started
(c) ${v_{\max }} = a\omega $
==> $\omega = \frac{{{v_{\max }}}}{a} = \frac{{10}}{4}$
Now, $v = \omega \sqrt {{a^2} - {y^2}} $
==> ${v^2} = {\omega ^2}({a^2} - {y^2})$
==> ${y^2} = {a^2} - \frac{{{v^2}}}{{{\omega ^2}}}$
$y = \sqrt{{a^2} - \frac{{{v^2}}}{{{\omega ^2}}}}$
$= \sqrt{{4^2} - \frac{5^2}{{({\frac{10}{4}})^2}}}$
$ = 2\sqrt 3 \,cm$
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
- 1An object of mass $0.5\, {kg}$ is executing simple harmonic motion. Its amplitude is $5\, {cm}$ and time period (T) is $0.2\, {s} .$ What will be the potential energy of the object at an instant $t=\frac{T}{4}$ s starting from mean position. Assume that the initial phase of the oscillation is zero. (In ${J}$)View Solution
- 2View SolutionA shooting game involves using a gun that fires by itself at random times. The player can only point the gun in a fixed direction while the target moves from side to side with simple harmonic motion, as shown. At which of the given region should the player aim in order to score the greatest number of hits?
- 3A function is represented by equationView Solution
$y = A\,\cos \,\omega t\,\cos \,2\omega t + A\,\sin \,\omega t\,\sin \,2\omega t$.
Than the nature of the function is
- 4Two simple pendulum first of bob mass $M_1$ and length $L_1$ second of bob mass $M_2$ and length $L_2$. $M_1 = M_2$ and $L_1 = 2L_2$. If these vibrational energy of both is same. Then which is correctView Solution
- 5A particle of mass $m$ moves in a one-dimensional potential energy $U(x) = -ax^2 + bx^4,$ where $'a'$ and $'b'$ are positive constants. The angular frequency of small oscillations about the minima of the potential energy is equal toView Solution
- 6Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to $A$ and $T,$ respectively. At time $t=0$ one particle has displacement $A$ while the other one has displacement $\frac {-A}{2}$ and they are moving towards each other. If they cross each other at time $t,$ then $t$ isView Solution
- 7If $x=5 \sin \left(\pi t+\frac{\pi}{3}\right) \mathrm{m}$ represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, areView Solution
- 8Two bodies $M$ and $N $ of equal masses are suspended from two separate massless springs of force constants $k_1$ and $k_2$ respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude $M$ to that of $N$ isView Solution
- 9Two particles are in $SHM$ in a straight line. Amplitude $A$ and time period $T$ of both the particles are equal. At time $t=0,$ one particle is at displacement $y_1= +A$ and the other at $y_2= -A/2,$ and they are approaching towards each other. After what time they cross each other ?View Solution
- 10A plank with a small block on top of it is under going vertical $SHM$ . Its period is $2\ sec$ . The minimum amplitude at which the block will separate from plank isView Solution