A particle of mass $$0.8$$ kg is executing simple harmonic motion with an amplitude of $$1.0$$ metre and periodic time $$11/7$$ $$sec$$. The kinetic energy (in$$ $$J) of the particle at the moment when it displacement is $$0.6$$ metre is $$________$$.

Question

Infinity Learn · Accepted Answer

We know that at a displacement y from mean position particle's velocity is given as

$v = ω \sqrt{(A^{2} - y^{2})}$

$or v = \frac{2 π}{T} \sqrt{(A^{2} - y^{2})}$

$= \frac{2 \times 3 .14}{(11 / 7)} \sqrt{[(1 .0)^{2} - (0 .6)^{2}]} = 3 .2 m / s$

Kinetic energy at this displacement is given by

$K = \frac{1}{2} {mv}^{2} = \frac{1}{2} \times 0 .8 \times (3 .2)^{2} = 4 .1 J$

Simple harmonic motion

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A particle of mass 0.8 kg is executing simple harmonic motion with an amplitude of 1.0 metre and periodic time 11/7 sec. The kinetic energy (in J) of the particle at the moment when it displacement is 0.6 metre is ________.

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Simple harmonic motion

Remember concepts with our Masterclasses.

A particle of mass 0.8 kg is executing simple harmonic motion with an amplitude of 1.0 metre and periodic time 11/7 sec. The kinetic energy (in J) of the particle at the moment when it displacement is 0.6 metre is ________.

We know that at a displacement y from mean position particle's velocity is given asv=ωA2−y2 or v=2πTA2−y2=2×3.14(11/7)(1.0)2−(0.6)2=3.2m/sKinetic energy at this displacement is given byK=12mv2=12×0.8×(3.2)2=4.1J

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