First slide

Simple hormonic motion

Question

A linear harmonic oscillator of forte constant 2x10⁶ N/m and amplitude of 0 . 01 m has a total mechanical energy of 160 J. Its

Easy

A

maximum PE. is 100 J
B

maximum K.E. is 100 J
C

maximum PE. is 160 J
D

minimum PE. is zero

Solution

Elastic potential energy, $= \frac{1}{2} {kx}^{2}$
$= \frac{1}{2} \times (2 \times 10^{6}) \times (0 \cdot 01)^{2} = 100 J TLe enerry is converted into kinetic onergy . . Maximum K . E . = 100 I Given that total mechanical energy is 16 O j Hence maximum potential enerry is 160 j So, b and c are correct .$

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