The amplitude of a particle executing $SHM$ about $O$ is $10\, cm.$ Then :
Advanced
Download our app for free and get started
$K.E.$ $=0.64 \mathrm{KE}_{\max }$
$v=0.8 v_{\max }$
$\therefore x=0.6 A=6 \mathrm{cm}$
$x=\frac{A}{2} P . E .=\frac{P E_{\max }}{4} K E=\frac{3}{4} P E_{\max }$
$\mathrm{KE}_{\max }=\mathrm{TE}$ at mean position
$x=\frac{A}{2} v=\frac{\sqrt{3} v_{\max }}{2}$
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
- 1The displacement of a particle executing SHM is given by $x=10 \sin \left(\omega t+\frac{\pi}{3}\right) \mathrm{m}$. The time period of motion is $3.14 \mathrm{~s}$. The velocity of the particle at $\mathrm{t}=0$is_________. $\mathrm{m} / \mathrm{s}$.View Solution
- 2A particle is performing simple harmonic motion with amplitude A and angular velocity ${\omega }$. The ratio of maximum velocity to maximum acceleration isView Solution
- 3A steady force of $120\ N$ is required to push a boat of mass $700\ kg$ through water at a constant speed of $1\ m/s$ . If the boat is fastened by a spring and held at $2\ m$ from the equilibrium position by a force of $450\ N$ , find the angular frequency of damped $SHM$ ..... $rad/s$View Solution
- 4A large horizontal surface moves up and down in $SHM$ with an amplitude of $1 \,cm$. If a mass of $10\, kg$ (which is placed on the surface) is to remain continually in contact with it, the maximum frequency of $S.H.M.$ will be ... $Hz$View Solution
- 5The displacement equation of a particle is $x = 3\sin 2t + 4\cos 2t.$ The amplitude and maximum velocity will be respectivelyView Solution
- 6A 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
- 7A mass $m = 1.0\,kg$ is put on a flat pan attached to a vertical spring fixed on the ground. The mass of the spring and the pan is negligible. When pressed slightly and released, the mass executes simple harmonic motion. The spring constant is $500\,N/m.$ What is the amplitude $A$ of the motion, so that the mass $m$ tends to get detached from the pan ? (Take $g = 10\,m/s^2$ ). The spring is stiff enough so that it does not get distorted during the motion.View Solution
- 8A particle is executing simple harmonic motion with a period of $T$ seconds and amplitude a metre. The shortest time it takes to reach a point $\frac{a}{{\sqrt 2 }}\,m$ from its mean position in seconds isView Solution
- 9When a mass $m$ is hung from the lower end of a spring of neglibgible mass, an extension $x$ is produced in the spring. The time period of oscillation isView Solution
- 10A $3\ kg$ sphere dropped through air has a terminal speed of $25\ m/s$. (Assume that the drag force is $-bv$.) Now suppose the sphere is attached to a spring of force constant $k = 300\ N/m$, and that it oscillates with an initial amplitude of $20\ cm$. What is the angular frequencu of its damped $SHM$? ..... $rad/s$View Solution