The potential energy of a particle executing S.H.M. is $ 2.5\, J$, when its displacement is half of amplitude. The total energy of the particle be .... $J$
A$18$
B$10$
C$12$
D$2.5$
Medium
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B$10$
b (b) $\frac{{{\rm{Potential\, energy (}}U{\rm{)}}}}{{{\rm{Total \,energy (}}E{\rm{)}}}} = \frac{{\frac{1}{2}m{\omega ^2}{y^2}}}{{\frac{1}{2}m{\omega ^2}{a^2}}} = \frac{{{y^2}}}{{{a^2}}}$
So $\frac{{2.5}}{E} = \frac{{{{\left( {\frac{a}{2}} \right)}^2}}}{{{a^2}}}$
==> $E = 10J$
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Two bodies of masses $1\, kg$ and $4\, kg$ are connected to a vertical spring, as shown in the figure. The smaller mass executes simple harmonic motion of angular frequency $25\, rad/s$, and amplitude $1.6\, cm$ while the bigger mass remains stationary on the ground. The maximum force exerted by the system on the floor is ..... $N$ ( take $g = 10\, ms^{-2}$)
A 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.