Question

Average kinetic energy with respect to time in shm

Answer 1

Answer:

purely kinetic. At the mean position, the velocity of the particle in S.H.M. is maximum and displacement is minimum, that is, x=0. Therefore, P.E. =1/2 K x2 = 0 and K.E. = 1/2 k ( a2 – x2) = 1/2 k ( a2 – o2) = 1/2 ka2. Thus, the total energy in simple harmonic motion is purely kinetic.

Answer 2

Answer:

The average kinetic energy with respect to time in simple harmonic motion is ${\dfrac{1}{4} ma^{2}\omega^{2}$.

Explanation:

Simple harmonic motion is the motion where the restoring force is directly proportional to the displacement of the body from its mean position.

The displacement $x$ of a particle in simple harmonic motion is given by

$x=asin\omega t$

Then the velocity is given by

$v=\dfrac{dx}{dt} =\dfrac{d(asin\omega t)}{dt}=a\omega cos\omega t$

Kinetic energy is given by

$KE=\dfrac{1}{2} mv^{2} =\dfrac{1}{2} m(a\omega cos\omega t)^{2} =\dfrac{1}{2} ma^{2}\omega^{2} cos^{2}\omega t$

Integrating KE to get the average kinetic energy with respect to time over the time period 0 to T,

$KE=\dfrac{\int\limits^T_0 {K.E.dt} }{\int\limits^T_0 dt}}$

     $=\dfrac{\int\limits^T_0 {\dfrac{1}{2} ma^{2}\omega^{2} cos^{2}\omega tdt} }{\int\limits^T_0dt }} =\dfrac{{\dfrac{1}{2} ma^{2}\omega^{2}\int\limits^T_0 cos^{2}\omega tdt} }{t\Big|\limits^T_0 }}$

    $=\dfrac{{\dfrac{1}{2} ma^{2}\omega^{2}{\displaystyle\int\limits^T_0 \dfrac{cos{2}\omega t+1}{2} dt} }}{T-0 }} =\dfrac{{\dfrac{1}{4} ma^{2}\omega^{2}\bigg({\int\limits^T_0 {cos{2}\omega t+{\int\limits^T_0} dt}\bigg)}}}{T }}$

    $=\dfrac{{\dfrac{1}{4} ma^{2}\omega^{2}\bigg({\dfrac{sin{2}\omega t}{2} {\bigg|\limits^T_0+{t\bigg|\limits^T_0}\bigg)}}}}{T }}=\dfrac{{\dfrac{1}{4} ma^{2}\omega^{2}\big(0 +{T}\big)}}{T }}={\dfrac{1}{4} ma^{2}\omega^{2}$

Therefore, the average kinetic energy with respect to time in simple harmonic motion is ${\dfrac{1}{4} ma^{2}\omega^{2}$.

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