The potential energy of a simple harmonic oscillator at mean position is $2\,joules$. If its mean $K.E.$ is $4\,joules$, its total energy will be .... $J$
A$7$
B$8$
C$10$
D$11$
Medium
Download our app for free and get started
C$10$
c Total energy $=\frac{1}{2} \mathrm{KA}^{2}+\mathrm{U}_{0}$
$\mathrm{U}_{0}=2 \mathrm{J}(\text { given })$ and $\frac{1}{4} \mathrm{KA}^{2}=4 \mathrm{J}$
So total energy $=8+2=10 \mathrm{J}$
Download our app
and 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.*
The potential energy of a particle of mass $1\, kg$ in motion along the $x-$ axis is given by $U = 4\,(1 -cos\,2x)$, where $x$ is in $metres$ . The period of small oscillation (in $second$ ) is
The maximum speed of a particle executing $S.H.M.$ is $1\,m/s$ and its maximum acceleration is $1.57\,m/se{c^2}$. The time period of the particle will be .... $\sec$
A particle is performing simple harmonic motion along $x$ -axis with amplitude $4\, cm$ and time period $1.2\, sec$. The minimum time taken by the particle to move from $x = 2\, cm$ to $x =+ 4 \,cm$ and back again is given by .... $s$
A block of mass $200\, g$ executing $SHM$ under the influence of a spring of spring constant $K=90\, N\,m^{-1}$ and a damping constant $b=40\, g\,s^{-1}$. The time elapsed for its amplitude to drop to half of its initial value is ...... $s$ (Given $ln\,\frac{1}{2} = -0.693$)
Two simple pendulums of length $1\, m$ and $4\, m$ respectively are both given small displacement in the same direction at the same instant. They will be again in phase after the shorter pendulum has completed number of oscillations equal to
A particle is performing $SHM$ according to the equation $x = (3\, cm)$ $\sin \,\left( {\frac{{2\pi t}}{{18}} + \frac{\pi }{6}} \right)$ where $t$ is in seconds. The distance travelled by the particle in $36\, s$ is ..... $cm$