Question

a mass of 0.5 kg oscillates harmonically with amplitude of 0.05m. the force constant of spring is 10N/m. calculate its period of oscillation and the maximum kinetic energy of the mass.​

Answer 1

Answer:

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Answer 2

Period of oscillation and the maximum kinetic energy of the mass is 19.87 s and 0.0125 J respectively.

Given :

Mass of oscillating body , m = 0.5 kg.

Amplitude, A = 0.05 m.

Spring constant , k = 10 N/m.

We know , Time period of a body oscillating harmonically is :

                                       $T=2\times \pi\times \sqrt{\dfrac{m}{k}}$

Putting value of m and k in above equation.

We get ,

$T=2\times \pi\times \sqrt{\dfrac{0.5}{0.05}}=19.87\ s.$

Kinetic energy is given by :

$K.E=\dfrac{1}{2}\times m\times (A-x)^2\times \omega^2$

For maximum kinetic energy x = 0 .

$K.E=\dfrac{1}{2}\times m\times A^2\times \omega^2$

Here, $\omega=\sqrt{\dfrac{k}{m}}=\sqrt{\dfrac{10}{0.5}}=4.47\ rad/s.$

Putting value of $\omega$ and k in above equation.

We get ,

$K.E=\dfrac{1}{2}\times 0.5\times 0.05^2\times 4.47^2= 0.0125 \ J.$

Hence, this is the required solution.

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