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A damped harmonic oscillator consists of a mass on a spring, with a damping force proportional to the speed of the block. If the spring constant is 350 N/m, the mass of the block is 240 g, the damping constant is 0.41 kg/s, and the block is displaced 7.5 cm from its equilibrium position and then released, what is its kinetic energy after one cycle?

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5 votes

Answer:


KE=0.7341\ J

Step-by-step explanation:

Given:

  • spring constant,
    k=350\ N.m^(-1)
  • mass of the block attached,
    m=0.24\ kg
  • damping constant,
    b=0.41\ kg.s^(-1)
  • amplitude of oscillation,
    A=0.075\ m

Now the frequency of damped oscillation is given as:


\omega'=\sqrt{(k)/(m) -(b^2)/(4m^2) }


\omega'=\sqrt{(350)/(0.24) -(0.41^2)/(4* 0.24^2) }


\omega'=38.1786\ rad.s^(-1)

Now time period of one oscillation:


T=(2\pi)/(\omega')


T=(2\pi)/(38.1786)


T=0.1646\ s

We know the equation of motion for the damped harmonic motion of a mass attached to a spring is given as:


x=A.e^(-(b.t/2m)).cos\ \omega'.t ........................(1)

From above we find the position of the mass:


x=0.075* e^(-(0.41* 0.1646)/(2* 0.24))* cos\ (38.1786* 0.1646)


x=0.06477\ m is the position after 1 cycle.

So, the Kinetic energy can be given as:


KE=(1)/(2) k.x^2


KE=0.5* 350* 0.06477^2


KE=0.7341\ J

User Shlomi Haver
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