(A) Given: Mass $m = 1 \ kg$,spring constant $k = 100 \ N \ m^{-1}$,amplitude $A = 10 \ cm = 0.1 \ m$,displacement $x = 5 \ cm = 0.05 \ m$.Angular fre

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(A) Given: Mass $m = 1 \ kg$,spring constant $k = 100 \ N \ m^{-1}$,amplitude $A = 10 \ cm = 0.1 \ m$,displacement $x = 5 \ cm = 0.05 \ m$.Angular frequency $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{100}{1}} = 10 \ rad \ s^{-1}$.Velocity at displacement $x$ is given by $v = \omega \sqrt{A^2 - x^2} = 10 \sqrt{0.1^2 - 0.05^2} = 10 \sqrt{0.01 - 0.0025} = 10 \sqrt{0.0075} = 10 \times 0.0866 = 0.866 \ m \ s^{-1}$.Kinetic energy $K.E. = \frac{1}{2} m v^2 = \frac{1}{2} \times 1 \times (0.866)^2 = 0.5 \times 0.75 = 0.375 \ J$.Potential energy $P.E. = \frac{1}{2} k x^2 = \frac{1}{2} \times 100 \times (0.05)^2 = 50 \times 0.0025 = 0.125 \ J$.

Class 11 Physics Oscillations SHM of Spring Mass System

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$A$ block of mass $1 \ kg$ is fastened to a spring of spring constant $100 \ N \ m^{-1}$. The block is pulled to a distance $x = 10 \ cm$ from its equilibrium position $(x = 0 \ cm)$ on a frictionless surface,from rest at $t = 0$. The kinetic energy and the potential energy of the block when it is $5 \ cm$ away from the mean position are:

AP EAMCET 2021 All AP EAMCET

A
$0.375 \ J, 0.125 \ J$
B
$0.125 \ J, 0.375 \ J$
C
$0.125 \ J, 0.125 \ J$
D
$0.375 \ J, 0.375 \ J$

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SHM of Spring Mass System

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DifficultJEE MAIN 2022

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MediumAIEEE 2007

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MediumMHT CET 2021

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The potential energy of a particle of mass $4 \, kg$ in motion along the $x$-axis is given by $U = 4(1 - \cos 4x) \, J$. The time period of the particle for small oscillations $(\sin \theta \simeq \theta)$ is $\left(\frac{\pi}{K}\right) \, s$. The value of $K$ is .......

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