Question

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?

Answer 1

`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`